Can we find these log expressions in terms of $\log_9 x$?

Can we find these log expressions in terms of $\log_9 x$?

Question

Given that $u=\log_9 x$, find, in terms of $u$:

1. $x$,

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1. $\log_9(3x)$,

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1. $\log_x 81$.

(Answers may not contain the word $\log$.)

Solution

Given that $u=\log_9 x$, find, in terms of $u$

1. $x$

By definition, we have that $x=9^u$.

Using what we know about the logarithm of a product, we get that $\log_9(3x)=\log_9 3+\log_9 x=\frac{1}{2}+u.$

Let’s start by writing $v=\log_x 81$.

By definition, we see that $81=x^v$ and so using part (a), $81=(9^u)^v$.

Since $9^2=81$, we must have that $uv=2$, or $v=2/u$. Therefore, $\log_x 81=2/u$.