# Solve the following equation

• May 10th 2009, 08:50 PM
Hapa
Solve the following equation
Having trouble getting started on a few of these.

1. 5^(x-10) = 125^x

2. (3/4) log x ^(3/4) + 9 = 0

thanks again!
• May 11th 2009, 09:03 AM
inzaghina
Hello..

1) 125^x=5^(3x) => x-10=3x => x=-5

2) 3/4 * log x ^(3/4) + 9 = 0 => 3/4*3/4 * log x = -9 => 1/16 * log x = -1 => log x = -16
if log = ln => x=e^(-16)
• May 11th 2009, 09:13 AM
Soroban
Hello, Hapa!

Quote:

$\displaystyle 1)\;\;5^{x-10} \:=\: 125^x$
We have: .$\displaystyle 5^{x-10} \;=\;\left(5^3\right)^x \quad\Rightarrow\quad 5^{x-10} \:=\:5^{3x}$

Equate exponents: .$\displaystyle x-10 \:=\:3x\quad\hdots\;\text{ etc.}$

Quote:

$\displaystyle 2)\;\;\tfrac{3}{4}\log\left(x ^{\frac{3}{4}}\right) + 9 \:=\: 0$

We have: .$\displaystyle \tfrac{3}{4}\log\left(x^{\frac{3}{4}}\right) \:=\:-9 \quad\Rightarrow\quad \tfrac{3}{4}\cdot\tfrac{3}{4}\log(x) \:=\:-9 \quad\Rightarrow\quad \tfrac{9}{16}\log(x) \:=\:-9$

. . . . . . $\displaystyle \log(x) \:=\:-16 \quad\Rightarrow\quad x \:=\:10^{-16}$

Edit: Too slow ... again!
• May 11th 2009, 09:43 AM
inzaghina
soroban, how do you do to write the equations so beautiful ? :)
• May 11th 2009, 04:11 PM
lmasud
he used mathtype to write it like that
and here is another way to solve:

5^(x-10)= 125^x

5^(x-10)= 5^(3x)

Take "log" of both sides:

(x-10)log5= (3x)log5

dividing both sides by log5:

x-10= 3x

x= -5