Can someone help me solve logarithmic equations?

Question

I really don't understand it.
For example:

solve
log3 (x - 3) = log9 x
(the first three and 9 are subscripts of the log)

Answer 1

You need to rewrite equation in exponential form:

9 ^ log base 3 (x-3) = x

Now here comes the fun part, simplifying the left side of that equation, which requires you to apply the inverse property of logs and exponents. That is, B^ (log base B x) = x

But 3 and 9 are not the same bases, yet you can write 9 as 3^2, which yields:

(3^2)^(log base 3 (x-3))
= 3^(2log base 3 (x-3))
= 3^ (log base 3 (x-3)²)
= (x-3)²

So now you have
(x-3)² = x
x² - 6x + 9 = x
x² - 7x +9 = 0
x = (7+- sqrt 13) / 2

But our only valid solution is x = (7+ sqrt 13) / 2 because the other causes a problem in the original equation, which is that you would be taking the log of a negative number.

Answer 2

log9 x = log3 x / log3 9 = log3 x / log3 (3^2) = (log3 x) / 2.
So we have log3 (x-3) = (1/2) log3 x = log3 (√x)
=> x-3 = √x.
=> (x-3)^2 = x
=> x^2 - 6x + 9 = x
=> x^2 - 7x + 9 = 0
=> x = (7 ± √(49-4.1.9)) / 2
= (7 ± √13) / 2.

Another way to solve this is to say
log3 (x-3) = y = log9 x
=> 3^y = x-3 and 9^y = x
=> 3^y = x-3 and (3^y)^2 = x
=> (x-3)^2 = x
and continue on as before.

Answer 3

log3 (x - 3) = log9 x
log3 (x - 3) = log3(x)/log3(9); using change of base formula: loga b=logc b/logc a

log3 (x - 3) = log3(x)/log3(3^2)
log3 (x - 3) = log3(x)/2
2log3 (x - 3) = log3(x)
log3 ((x - 3)^2) = log3(x)
(x - 3)^2 = (x)
x^2-6x+9=x
x^2-7x+9=0