Question #93891

1 Answer
Jan 24, 2018

x=2

Explanation:

Your notation is very ambiguous. I am assuming the equation is
3^(2x)-2*3^(x+2)+81=0. If not, then this has all been in vain.
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3^(2x)-2*3^(x+2)+81=0

We can rewrite this using the laws of indices:

3^(2x)=(3^x)^2

-2*3^(x+2)=-2*3^x*3^2

So we have:

(3^x)^2-2*3^x*3^2+81=0

Simplify:

(3^x)^2-18*3^x+81=0

This is a quadratic in 3^x

Let u=3^x

Then:

u^2-18u+81=0

Factor:

(u-9)^2=0=>u=9

But:

u=3^x=>3^x=9

3^x=3^2=>x=2