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Chapter 3: Problem 23

Evaluate the indicated expression. Do not use a calculator for these exercises. $$ \log _{16} 32 $$

Short Answer

Expert verified
The short answer to the given problem is \(\log_{16} 32 = \frac{5}{4}\).

Step by step solution

01

Rewrite the expression as an exponential equation

We start by rewriting the given log expression as an exponential equation using the definition of logarithms. If \(y = \log_b{x}\), then \(b^y = x\). In this case, \(y = \log_{16}{32}\) and the base \(b = 16\). We write it as an exponential equation: $$ 16^y = 32 $$
02

Rewrite both sides with a common base

Both 16 and 32 are powers of 2, so we rewrite them as such: $$ (2^4)^y = 2^5 $$
03

Use the power rules to simplify

Now, we can use the power rule \((a^{m})^{n} = a^{mn}\) to simplify the expression: $$ 2^{4y} = 2^5 $$
04

Match the exponents

Since both sides of the equation have the same base, we can match the exponents to find the value of \(y\): $$ 4y = 5 $$
05

Solve for y

Divide both sides of the equation by 4 to find the value of \(y\): $$ y = \frac{5}{4} $$
06

Final Answer

So, \(\log_{16} 32 = \frac{5}{4}\).

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Most popular questions from this chapter

Show that $$ (\cosh x+\sinh x)^{t}=\cosh (t x)+\sinh (t x) $$ for all real numbers \(x\) and \(t\).

For each of the functions \(f\); (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(f^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part ( \(c\) ) by verifying that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I .\) (Recall that \(I\) is the function defined by \(I(x)=x .)\) \(f(x)=5 e^{9 x}\)

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Show that \(\sinh\) is a one-to-one function and that its inverse is given by the formula $$ (\sinh )^{-1}(y)=\ln \left(y+\sqrt{y^{2}+1}\right) $$ for every real number \(y\).
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