1. ## Solve for x

how do you solve

3^(2x)-3^(x+2)+8=0

2. i think this needs a change of base on the 8

8 = 3^log.3. 8

in general

a^x = b^(log.b. a)(x)

move the term with the negative to the other side so they are all positive
then since you have all the numbers with base 3 you can say the bases are equal therefore the exponents must be equal

and then you can equate the exponents
i get x = 2 - log.3. 8

3. Originally Posted by fatman
how do you solve

3^(2x)-3^(x+2)+8=0
lets rewrite this as

$(3^x)^2-3^23^x+8=0$

this is quadratic in $3^x$ so to simplify let $u=3^x$

The we get

$u^2-9u+8=0 \iff (u-1)(u-8)=0$ so u =1 and u=8

but $u=3^x$ so we get

$1=3^x \iff \log_3(1)= \log_33^x \iff 0=x$

$8=3^x \iff \ln(8)=\ln(3^x) \iff \ln(8)=x\ln(3) \iff \frac{\ln(8)}{\ln(3)}=x$

Good luck.

4. oh i think my mistake was when i said that since the exponents of
A and B and C are all the same in the relation:

A+B=C

when in fact it only works when you have a relation:
A=B

anyone confirm?

5. my mistake, i made a blunder with the exponent rules

6. ## Thanks

Thank you very much for your all your help with quick replies