# Thread: Basic indices question

1. ## 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.

2. ## Re: Basic indices question

First of all,
If

$\displaystyle a^x = b$

then

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

so if

$\displaystyle 2^2 = 4$

then

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

Now for your problem:

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

Raise both sides by the power

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

Which gives,

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

Since we know that

$\displaystyle 3^2 = 9$

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

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

$\displaystyle x = 2y - 2$

Hope this helps!

3. ## 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: $\displaystyle \log_a(a^x)=x \quad a^{\log_a(x)}=x$

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

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

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

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

$\displaystyle x=2(y-1)$

$\displaystyle x=2y-2$