• Jun 19th 2006, 08:08 AM
kbryant05
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!! :eek:

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
• Jun 19th 2006, 08:18 AM
Jameson
1. $\sqrt{36}=6$
$\log_{36}6=\frac{1}{2}$

2. $\ln(e^{5x})=5x\ln(e)=5x(1)=5x$
• Jun 19th 2006, 08:40 AM
Soroban
Hello, kbryant05!

Quote:

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

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

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

Therefore: . $x = 6$

Quote:

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

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