Intro to Logarithms

Section 13.4 Intro to Logarithms

Logarithms are a kind of inverse of exponents.

Example 13.4.1. Motivation for Logs with Base 10.

Consider these powers of 10,

\begin{align*} 10^1 \amp = 10\\ 10^2 \amp = 100\\ 10^3 \amp = 1000\\ 10^4 \amp = 10000 \end{align*}

These tell you that 10 to an exponent gives a number. Now reverse the question. Instead of asking what \(10^3\) equals, ask: 10 to what power equals 1000? Since \(10^3=1000\text{,}\) the answer is 3.

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Logarithms are a shorthand for this reverse question. The expression \(\log_{10}(x)\) (read this as “log base 10 of \(x\)”) means the exponent you put on 10 to get \(x\text{.}\) Intuitively, \(\log_{10}{x}\) is the number of 10s you need to multiply together in order to get \(x\text{.}\) So, for example,

\begin{align*} \begin{array}{lcl} \amp \text{means...} \amp \\ 10^3=1000 \amp \longrightarrow \amp \log_{10}(1000)=3 \\ 10^5=100000 \amp \longrightarrow \amp \log_{10}(100000)=5 \\ 10^0=1 \amp \longrightarrow \amp \log_{10}(1)=0 \end{array} \end{align*}

In other words,

Powers convert exponents to numbers

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Logs convert numbers to exponent.

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In this way, logarithms are a kind of inverse of exponents.

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Example 13.4.2. Approximating Logarithms.

We know that for the equation \(2^x=8\text{,}\) the answer is \(x=3\text{,}\) because \(2^3=8\text{.}\) However, what about the equation \(2^x=11\text{?}\)

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We know that \(2^3=8\) and \(2^4=16\text{,}\) so we expect the solution \(x\) to be between 3 and 4. Maybe a better guess is in the middle, like \(x=3.5\text{.}\) Here, \(2^{3.5} \approx 11.3\text{,}\) which is a bit more than 11. Try \(x=3.4\text{,}\) and we get \(2^{3.4} \approx 10.6\text{,}\) which is a bit less than 11. So, the solution is between 3.4 and 3.5.

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Using logarithms, we can say that the exact answer is \(\log_{2}{11}\) (read this as “log base 2 of 11”), which means the exponent you must put on 2 to get 11. It turns out that \(\log_2{11} \approx 3.45943 \dots\text{.}\)

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You can try this on your calculator. You’ll see that \(2^{3.45943} = 10.99998766\text{,}\) which is very, very close to 11.

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Example 13.4.3. Logarithm Examples.

Similarly,

\(\log_{4} 64 = 3\text{,}\) because \(4^3=64\text{.}\)

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\(\log_{10}1000000 = 6\text{,}\) because \(10^6=1000000\text{.}\)

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\(\log_{1/2}\left(\frac{1}{8}\right)=3\text{,}\) because \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)

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Subsection 13.4.1 Logarithms

The logarithm base b of a, denoted by \(\log_{b}a\text{,}\) is the exponent that \(b\) must be raised to, to get \(a\text{.}\) In other words,

\begin{gather*} \boxed{\log_{b}a = c \quad \text{is equivalent to} \quad b^c = a} \end{gather*}

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The equations \(\log_{b}a = c\) (logarithmic form) and \(b^c = a\) (exponential form) are two ways of expressing the same relationship between the numbers \(a, b\text{,}\) and \(c\text{.}\)

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With the logarithmic form \(\log_{b}a = c\text{,}\)

\(b\) is called the base of the logarithm.

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\(a\) is called the argument (the value inputted into the function).

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\(c\) is the logarithm (since it equals the logarithmic expression).

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With the exponential form \(b^c = a\text{,}\)

\(b\) is also the base.

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\(c\) is the exponent.

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\(a\) is the result of the power.

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Notice that the exponent \(c\) is equivalent to the logarithm above (since logarithms are an exponent), and that the result \(a\) is equivalent to the argument above.

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Exercise Group 13.4.1. Converting Forms.

Convert each logarithmic or exponential equation to exponential or logarithmic form.

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(a)

\(\log 0.1=-1\)

Answer.

\(10^{-1}=0.1\)

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(b)

\(\log_{5}3125=5\)

Answer.

\(5^5=3125\)

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(c)

\(8^0=1\)

Answer.

\(\log_{8}{1}=0\)

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(d)

\(8^{\frac{2}{3}}=4\)

Answer.

\(\log_{8}{4}=\frac{2}{3}\)

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(e)

\(9^{-1}=\frac{1}{9}\)

Answer.

\(\log_{9}\left(\frac{1}{9}\right)=-1\)

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(f)

\(\log_3{729}=6\)

Answer.

\(3^6=729\)

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(g)

\(\log_{\sqrt{3}}{3}=2\)

Answer.

\(\left(\sqrt{3}\right)^2=3\)

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(h)

\(3^{-2}=\frac{1}{9}\)

Answer.

\(\log_{3}\left(\frac{1}{9}\right)=-2\)

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(i)

\(\log_{2x}{(x+1)}=3\)

Answer.

\((2x)^3=x+1\)

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Logarithms are a name for a certain exponent. In particular, \(\log_b{a}\) is the exponent to which \(b\) must be raised, to obtain \(a\text{.}\)

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When you see “\(\log{\text{something}}\)”, mentally translate into “the exponent you’d put on the base to get what’s inside”.

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Exercise Group 13.4.2. Evaluating Basic Logarithms.

Evaluate each logarithm.

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(a)

\(\log_{4}16\)

Answer.

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(b)

\(\log_{2}32\)

Answer.

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(c)

\(\log_{3}9\)

Answer.

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(d)

\(\log 10000\)

Answer.

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(e)

\(\log_3{81}\)

Answer.

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(f)

\(\log_{7}49\)

Answer.

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(g)

\(\log_{5}3125\)

Answer.

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(h)

\(\log 1000000\)

Answer.

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(i)

\(\log_{4}256\)

Answer.

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(j)

\(\log_{3}729\)

Answer.

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Subsection 13.4.2 Logarithm Bases

The two most commonly-used logarithm bases are:

Base 10, denoted by \(\log_{10}\text{,}\) called the common logarithm, which is used in many applied sciences.

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Base \(e\text{,}\) called the natural logarithm, which is often used in calculus.

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These are often denoted by simply \(\log\) and \(\ln\text{,}\) respectively. In other words,

\begin{gather*} \log{x} = \log_{10}{x} \quad \text{and} \quad \ln{x} = \log_e{x} \end{gather*}

These two logarithms are included on most calculators. The natural logarithm is denoted by \(\ln\) because it is an abbreviation of the Latin term “logarithmus naturali”.

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Remark 13.4.4.

In computer science, often log base 2 is used, or \(\log_2\text{.}\)

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Remark 13.4.5.

For logarithms, we use three letters \(\log\) to denote the common logarithm, and two letters \(\ln\) to denote the natural logarithm.

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Typically, functions have been represented by a single letter, such that \(f\) for \(f(x)\) or \(g\) for \(g(x)\text{.}\)

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In this way, \(\log{(x)}\) or \(\log_2{(x)}\) represents a function just like \(f(x)\) does. However, we often write logarithms without parentheses as \(\log{x}\) or \(\log_2{x}\text{.}\)

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If you have seen the trig functions like \(\sin{x}\) or \(\cos{x}\text{,}\) then \(\log{x}\) is a similar kind of notation: \(\log{x}\) means the log function with \(x\) as its input, just like \(\sin{x}\) means the sine function with \(x\) as its input. In particular, \(\log{10}\) does not mean “\(\log\) times 10”, just like \(\sin{30^{\circ}}\) doesn’t mean “\(\sin\) times 30”.

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If a log contains more than one term, then we use brackets. For example, \(\log(x + y)\text{.}\) Or, if there are a lot of factors, e.g. \(\log\left(\frac{x^2 y^3}{(x-1)^5}\right)\text{.}\)

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Subsection 13.4.3 Basic Properties of Logarithms

Example 13.4.6. Logarithm of the Base.

Consider \(\log_5{5}\text{.}\) This means the exponent to put on 5 to get 5. This is 1, since \(5^1=5\text{.}\) Similarly

\(\displaystyle \log_3{3} = 1\)

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\(\displaystyle \log_8{8} = 1\)

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\(\displaystyle \log{10} = 1\)

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Theorem 13.4.7. Logarithm of the base.

\begin{gather*} \boxed{\log_{b}b = 1} \qquad \text{} \end{gather*}

Because, the exponent you need to put on \(b\) to get \(b\) is 1.

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Example 13.4.8. Logarithm of One.

Consider \(\log_{3}{1}\text{.}\) This means the exponent to put on 3 to get 1. This is 0, because \(3^0=1\text{.}\) Similarly,

\(\displaystyle \log_8{1}=0\)

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\(\displaystyle \log_5{1}=0\)

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\(\displaystyle \log{1}=0\)

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Theorem 13.4.9. Logarithm of 1.

\begin{gather*} \boxed{\log_{b}1 = 0} \qquad \text{} \end{gather*}

Because, the exponent you need to put on \(b\) to get 1 is 0.

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Subsection 13.4.4 Evaluating Logarithms (By Hand)

Exercise Group 13.4.3. Evaluating Logarithms.

Evaluate each logarithm.

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(a)

\(\log_{7}\sqrt{7}\)

Answer.

\(\frac{1}{2}\)

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(b)

\(\log_{2}\left(\frac{1}{2}\right)\)

Answer.

-1

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(c)

\(\log_{\sqrt{2}}4\)

Answer.

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(d)

\(\log_{6}\left(\frac{1}{216}\right)\)

Answer.

-3

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(e)

\(\log{0.001}\)

Answer.

-3

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(f)

\(\log_{4}\left(\frac{1}{32}\right)\)

Answer.

\(-\frac{5}{2}\)

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(g)

\(\log_{3}\sqrt{3}\)

Answer.

\(\frac{1}{2}\)

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(h)

\(\log_2{\sqrt{32}}\)

Answer.

\(\frac{5}{2}\)

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(i)

\(\log_{8}\left(2\sqrt[3]{2}\right)\)

Answer.

\(\frac{4}{9}\)

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(j)

\(\log_{2}\left(\sqrt[3]{4}\right)\)

Answer.

\(\frac{2}{3}\)

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(k)

\(\log_{3}\left(3\sqrt{3}\right)\)

Answer.

\(\frac{3}{2}\)

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(l)

\(\log_{9}\left(3\sqrt{3}\right)\)

Answer.

\(\frac{3}{4}\)

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(m)

\(\log_{5}\left(\frac{1}{25}\right)\)

Answer.

-2

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(n)

\(\log_{2}\left(8\sqrt{2}\right)\)

Answer.

\(\frac{7}{2}\)

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(o)

\(\log_{1/2}16\)

Answer.

-4

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(p)

\(\log_{27}\left(3\sqrt{3}\right)\)

Answer.

\(\frac{1}{2}\)

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(q)

\(\log_{1/3}{27}\)

Answer.

-3

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(r)

\(\log_{2}\sqrt[6]{32}\)

Answer.

\(\frac{5}{6}\)

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(s)

\(\log_2(\log_3(\log_4{64}))\)

Answer.

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(t)

\(\log_4(\log_2(\log(10^{16})))\)

Answer.

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Subsection 13.4.5 Logs of Negative Numbers

It turns out, we can’t take the logarithm of a negative number, or 0.

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Example 13.4.10. Log of a Negative Number.

Consider \(\log_2{(-4)}\text{.}\) This number represents the exponent to put on 2 to get \(-4\text{.}\) However, the equation \(2^x=-4\) has no solution, because 2 to the power of any number will always result in a positive number, never a negative.

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Example 13.4.11. Log of Zero.

Consider \(\log_2{0}\text{.}\) This also doesn’t work, because \(2^x=0\) has no solution. Again, 2 to the power of any number will always be positive, and so will never equal to 0.

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In general,

\begin{gather*} \boxed{\text{You can only take the log of a positive number}} \end{gather*}

This is because for any base \(b\text{,}\) the expression \(\log_b{x}\) means the exponent needed to raise \(b\) to get \(x\text{.}\) Since \(b\) to the power of any number is always positive, there is only an answer if \(x\) is positive.

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Subsection 13.4.6 Evaluating Logarithms Using a Calculator

My recommended calculator Casio fx-991ES Plus C 2nd edition (or some other similar scientific calculators, as well as the TI-84 Plus CE) can calculate logarithms of any base.

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However, many older and weaker calculators only have buttons for:

Base 10 (common logarithms) denoted by the \(\log\) key

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Natural logarithms (base \(e\)) denoted by \(\ln\)

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In fact, base 10 logs are enough to evaluate logs of any base, because of this rule,

\begin{gather*} \boxed{\log_b{a} = \frac{\log{a}}{\log{b}}} \end{gather*}

This allows us to change a log of any base (\(b\)) to something involving logs base 10. It also works for natural logarithms, but you can just stick with logs base 10.

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Theorem 13.4.12. Change-of-base property.

\begin{gather*} \log_b{a} = \frac{\log{a}}{\log{b}} = \frac{\ln{a}}{\ln{b}} \qquad \qquad \text{for any } b \gt 0, b \neq 1 \end{gather*}

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Example 13.4.13. Using Change of Base.

For example,

\(\displaystyle \log_2{15} = \frac{\log{15}}{\log{2}} \approx 3.907\)

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\(\log_2{8} = \frac{\log{8}}{\log{2}} = 3\) (we could have found this one without a calculator)

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This equality is a particular case of what is called the change-of-base property, which we’ll explore further later on.

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Subsection 13.4.7 Inverse Properties of Logarithms

Example 13.4.14. Logarithm of a Power.

Consider \(\log_3(3^5)\text{.}\) This means the exponent that 3 must be raised, to get \(3^5\text{.}\) The answer is 5, because literally, 3 to the 5 is 3 to the 5. Similarly,

\(\displaystyle \log_5(5^8) = 8\)

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\(\displaystyle \log_8(8^{100})=100\)

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\(\displaystyle \log{10^4} = 4\)

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In general, if you take the log of a power of the base, the answer is the exponent.

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Theorem 13.4.15. Logarithm of a power of the base.

\begin{align*} \boxed{\log_{b}b^x = x} \amp\amp \text{} \end{align*}

Intuitively, \(\log_{b}(b^x)\) is the exponent to put on \(b\) to get \(b^x\text{,}\) which is literally \(x\text{.}\)

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Example 13.4.16. Power of a Logarithm.

Consider \(4^{\log_4{11}}\text{.}\) The exponent of \(\log_4{11}\) represents the exponent to put on 4 to get 11. We are taking 4 to the power of this number, so by definition, we should get 11. Therefore, \(4^{\log_4{11}}=11\text{.}\)

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Similarly,

\(\displaystyle 5^{\log_5{125}} = 125\)

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\(\displaystyle 3^{\log_3 \frac{1}{16}} = \frac{1}{16}\)

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\(\displaystyle 6^{\log_6{x}} = x\)

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Theorem 13.4.17. Power of a logarithm with the same base.

\begin{gather*} \boxed{b^{\log_{b}x} = x} \qquad \text{} \end{gather*}

Intuitively, \(\log_{b}x\) is the exponent which \(b\) must be raised to get \(x\text{.}\) So, if we raise \(b\) to that exponent, we get (literally) \(x\text{.}\)

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