Question

Now that we have defined $\ln x$ with an integral, we can define a general logarithm with base b as ${\log }_{b}x=\frac{1}{\ln b}\ln x$. In Exercises 24–26 you will investigate this definition of ${\log }_{b}x$.

We can now define general exponential functions $b^x$as the inverses of the general logarithmic functions ${\log }_{b}x$. What can you say about ${\log }_b{b}^y$. Use this information to simplify role="math" localid="1648736566259" $\frac{1}{\ln b}{\int }_1^{b^y}\frac{1}{t}dt$ so that it is written without an integral.

Short answer

The function, since they are inverses to each other, it will be equal to the exponent y, ${\log }_b{b}^y=y$.

And the value of the function $\frac{1}{\ln b}{\int }_1^{b^y}\frac{1}{t}dt=y$.

Step-by-step solution

Step 1. Given Information.

${\log }_{b}x=\frac{1}{\ln b}\ln x$

Step 2. Find logbby.

${\log }_b{b}^y=\frac{\log b^y}{\log b}$

Since they are inverses to each other, it will be equal to the exponent $y$.

${\log }_b{b}^y=y$

Step 3. Simplify 1ln b∫1by1tdt.

$$\begin{gathered} \frac{1}{\ln b}{\int }_1^{b^y}\frac{1}{t}dt=\frac{1}{\ln b}{[\ln x]}_1^{b^y} \\ =\frac{1}{\ln b}(\ln b^y-\ln 1) \\ =\frac{1}{\ln b}(\ln b^y-0) \\ =\frac{\ln b^y}{\ln b} \\ =y \end{gathered}$$