solve for x anatytically and exactly

[renegade05]

Alrighty, I have been working this out for like 2 hours and still nothing.

Question: \(\displaystyle 9^x-2(3^x)=15\)

I made \(\displaystyle 9^x = 3^{2x}\) to simplify things and to pick

\(\displaystyle \log_3\) as my base.

This is where i get stuck. I have tried moving the 15 around from side to side, i have tried to move the 2(3^x) around but i always get stuck.

I know you guys ask for peoples work before helping them. But i dont know where to begin with this one. I need guidance on the very first step.

Thanks!

 

[galactus]

Make a substitution.

Let \(\displaystyle u=3^{x}\)

Then, we have:

\(\displaystyle u^{2}-2u-15=0\)

solve the quadratic and resub. Ignore the negative result.

 

[renegade05]

I get:
\(\displaystyle \log_3(5)\)

Dam it, I hate when I dont see something so simple like that.

 

[BigGlenntheHeavy]

\(\displaystyle 9^x-2(3^x) \ = \ 15 \ \implies \ (3^2)^x-2(3^x)-15=0 \ \implies \ (3^x)^2-2(3^x)-15 \ = \ 0.\)

\(\displaystyle let \ u \ = \ 3^x, \ then \ u^2-2u-15 \ = \ (u-5)(u+3) \ = \ 0.\)

\(\displaystyle u \ = \ 5 \ \implies \ 5 \ = \ 3^x \ \implies \ x[ln(3)] \ = \ [ln(5)], \ hence \ x \ = \ \frac{ln(5)}{ln(3)}, \ QED.\)

 

[Denis]

\(\displaystyle \ QED.\)

Question Every Detail ?

 

[renegade05]

\(\displaystyle 9^x-2(3^x) \ = \ 15 \ \implies \ (3^2)^x-2(3^x)-15=0 \ \implies \ (3^x)^2-2(3^x)-15 \ = \ 0.\)

\(\displaystyle let \ u \ = \ 3^x, \ then \ u^2-2u-15 \ = \ (u-5)(u+3) \ = \ 0.\)

\(\displaystyle u \ = \ 5 \ \implies \ 5 \ = \ 3^x \ \implies \ x[ln(3)] \ = \ [ln(5)], \ hence \ x \ = \ \frac{ln(5)}{ln(3)}, \ QED.\)

same as:

I get:
\(\displaystyle \log_3(5)\)

right?

and ya what is QED?