[SOLVED] need logarithm help

**kelvinly** · March 17th 2010, 05:14 PM

hi i need help on solving these questions with all the log rules i'm confusing myself..

1). 7^(x-2) = 1/2^(x-2)

2). 10^(x^2+2x-8) = 1

3). (1/4)^(x+2) * 32^(x-1) = 64^(x-2)

i know i have to solve for x but i keep getting the wrong answers when i plug it into my graph and find the intersections for the answer.

thanks!

**Anonymous1** · March 17th 2010, 05:30 PM

For (1) do you mean $7^{x-2}= \frac{1}{2^{x-2}}?$

Then, $(x-2)\log(7) = (2-x)\log(2)$

$=> (x-2)=(2-x)\frac{\log(2)}{\log(7)}$

Let $\frac{\log(2)}{\log(7)} = \alpha$

and solve to find $x= 2$

For (2) take the log of both sides:

$\log_{10} (10^{(x^2+2x-8)}) = log_{10}(1)$

$=> x^2+2x-8 = \log_{10}(1)$

$=> x^2+2x-8-\log_{10}(1) = 0$

Now use the quadratic equation to find your solutions. Where,

$a= 1$
$b= 2$
$c= -(8+\log_{10}(1))$

For (3) again take the log of both sides.

$\log[(1/4)^{x+2}\times 32^{x-1}] = \log[64^{x-2}]$

$=> (x+2)\log(1/4)+(x-1)\log(32) = (x-2)\log(64)$

$...$

**kelvinly** · March 17th 2010, 06:39 PM

wow thanks a lot! it really helped, i was really off with some of them

**Anonymous1** · March 17th 2010, 06:47 PM

Logarithms with start to get intuitive, just keep practicing.

Good luck,
Anonymous

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