exponential base. logarithm.

ninjuhtime · #1 $Old$ November 30th, 2008, 08:34 PM

so working on a few exponential functions. i know that you can make the exponents equal to one another if the bases are the same but what if there is a single variable and the terms are multiplied/added.

what i have so far:

$3^{2x}-2*3^{x+5}+3^{10} = 0$
$(3^{2x})(3^{x+5})+3^{10} = 2$

is it correct to move over the 2 to work with the same bases easier?
or is this completely off and i should use logarithm and natural log? if so, please give hints as to how to work it out.

thankyou~

Soroban · #2 $Old$ November 30th, 2008, 09:14 PM

Hello, ninjuhtime!

Quote:

What i have so far:

$3^{2x}-2\cdot3^{x+5}+3^{10} \:= \:0$

$(3^{2x})(3^{x+5})+3^{10} = 2$ . . . . Definitely illegal!

It's hard to see it, but the expression is a perfect square . . .

We have: . $\left(3^x - 3^5\right)^2 \:=\:0 \quad\Rightarrow\quad 3^x - 3^5 \:=\:0 \quad\Rightarrow\quad 3^x \:=\:3^5 \quad\Rightarrow\quad\boxed{ x \:=\:5}$

ninjuhtime · #3 $Old$ November 30th, 2008, 10:15 PM

thankyou! i guess i oversaw what it was since long problems are usually intimidating.