# Basic indices question

• Oct 10th 2011, 01:26 PM
Jaxx
Basic indices question
First I'd like to say hi since this is my first post on the forum and it looks great and I look forward to talking maths with you guys in the future.

As for my question, I'm stumped on a little bit of rearranging and would like a bit of an explanation.

Given that 3^x = 9^(y-1), show that x = 2y -2.

• Oct 10th 2011, 01:42 PM
FrameOfMind
Re: Basic indices question
First of all,
If

$a^x = b$

then

$a = b^\frac {1}{x}$

so if

$2^2 = 4$

then

$2 = 4^\frac {1}{2}$

$3^x = 9^(^y^-^1^)$

Raise both sides by the power

$\frac{1}{(y-1)}$

Which gives,

$3^\frac {x}{(y-1)} = 9^\frac {(y-1)}{(y-1)} = 9^1 = 9$

Since we know that

$3^2 = 9$

$\frac {x}{y-1}$ must equal 2.

$\therefore\frac {x}{y-1} = 2$

$x = 2y - 2$

Hope this helps!
• Oct 11th 2011, 01:58 PM
skoker
Re: Basic indices question
to solve an exponential equation you use logarithms.
to solve an logarithmic equation you use exponentiation.

you have an exponential equation so take the log of both sided. you need to know the inverse property. and the one to one property. should be in your book.

inverse property: $\log_a(a^x)=x \quad a^{\log_a(x)}=x$

one to one property: $a^x=a^y \iff x=y$

$3^x=9^{y-1}$

$3^x=3^{2(y-1)}$

$\log_3(3^x)=\log_3(3^{2(y-1)})$

$x=2(y-1)$

$x=2y-2$