1. ## Logarithmic Help Please

I have a few problems I am having difficulty with if someone wouldn't mind helping me please. I am a little confused on how to show these types of problems. I can do a few but these ones are getting to me.
thank you!!

First problem:

Write in logarithmic from

square root sign only over the #36 then = 6

Second Problem:

simplify expression:

In e^5x

third problem:

solve equation:
3^x-2=81

& the fourth is the same to solve the equation:

log x= 1 over2
25

2. 1. $\displaystyle \sqrt{36}=6$
$\displaystyle \log_{36}6=\frac{1}{2}$

2. $\displaystyle \ln(e^{5x})=5x\ln(e)=5x(1)=5x$

3. Hello, kbryant05!

3) Solve: $\displaystyle 3^{x-2} \,= \,81$
With exponential equations, try to get the same base on both sides.

Since $\displaystyle 81 = 3^4$, we have: $\displaystyle 3^{x-2}\:=\:3^4$

The bases are equal, so the exponents are equal: .$\displaystyle x - 2\:=\:4$

Therefore: .$\displaystyle x = 6$

4) Solve: $\displaystyle \log_{25}x \,= \,\frac{1}{2}$
You should know how to rewrite this in exponential form: .$\displaystyle x \:=\:25^{\frac{1}{2}$

Therefore: .$\displaystyle x\:=\:\sqrt{25}\:=\:5$