Solve for $\displaystyle x$:$\displaystyle \log_{2008}\left(x-3\right)+\log_{2009}\left(x-3\right)=3-\lg\left(x^5-24\right)$
Well, we are given:
$\displaystyle log_{2008}{(x - 3)} + log_{2009}{(x - 3)} = 3 - log(x^5 - 24)$
Well, let's deal with the right side first, they are currently in different bases, therefore we must change the bases by using the change of base formula:
$\displaystyle log_a{b} = \frac{log{b}}{log{a}}$
Where we are using base 10 because the right side has a base 10 logarithm.
Therefore we get:
$\displaystyle \frac{1}{log{2008}}log{(x - 3)} + \frac{1}{log{2009}}log{(x - 3)} = 3 - log{(x^5 - 24)}$
Where:
$\displaystyle log{2008} = 3.203$
and
$\displaystyle log{2009} = 3.203$
(They are very close together, where the first difference is in the ten thousandths place.)
So we get:
$\displaystyle 0.31223log{(x - 3)} + 0.31223log{(x - 3)} = 0.31223(log{(x - 3)^2})$
$\displaystyle 0.62446log{(x - 3)} = 3 - log{(x^5 - 24)}$
There's really not an easy way to solve this, so I'm going to make the following approximation:
$\displaystyle 0.62446 \approx \frac{2}{3}$
$\displaystyle log{(x - 3)^{\frac{2}{3}}} = 3 - log(x^5 - 24)$
$\displaystyle log{(x - 3)^{\frac{2}{3}}} + log(x^5 - 24) = 3$
$\displaystyle log{(x - 3)^{\frac{2}{3}}(x^5 - 24)} = 3$
$\displaystyle (x - 3)^{\frac{2}{3}}(x^5 - 24) = 10^3$
$\displaystyle \sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2}$
$\displaystyle (x^5 - 24) = \frac{1000}{(x - 3)^5}$
$\displaystyle (x^5 - 24)(x + (-3))^5 = 1000$
$\displaystyle (x^5 - 24)(x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243) = 1000$
$\displaystyle x^{10} - 15x^9 + 90x^8 - 270x^7 + 405x^6 - 267x^5 + 360x^4 - 2160^3 + 6480x^2 - 9720x + 4832 = 0$
You get:
$\displaystyle x = 4$
To verify:
$\displaystyle
\frac{\log{1}}{log{2008}}+ \frac{\log{1}}{log{2009}} = 3 - \log\left(1000\right)
$
$\displaystyle 0 + 0 = 3 - 3 = 0$
And there you go.
All I did was cube both sides, and since cubing anything keeps the sign it currently has, I didn't need a plus or minus, but you are right, I was a power off:
$\displaystyle
\sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2}
$
$\displaystyle (\sqrt[3]{x^5 - 24})^3 = \left(\frac{10}{(x - 3)^2}\right)^3$
Evaluated, it equals:
$\displaystyle (x^5 - 24) = \frac{10^3}{((x - 3)^2)^3} = \frac{1000}{(x - 3)^6}$
But nonetheless, 4 is still the answer.
Just to clarify: $\displaystyle x=4$ is a solution. There can't be any other solutions because the LHS is strictly increasing but RHS is strictly decreasing because the fuction $\displaystyle \log_bx$ is stricly increasing if $\displaystyle b>1$ ($\displaystyle x>0$).
With equations have orders that high, there are very few differences. As a matter of fact, with the equation that I had before(the wrong one), it had all of its roots at (4,0).
Just for fun:
$\displaystyle (x + (-3))^6$
$\displaystyle = x^6 - 18x^5 + 135x^4 - 540x^3 + 1214x^2 - 1458x + 729$
Now if we take $\displaystyle (x^5 - 24)$ and multiply we get:
$\displaystyle x^{11} - 18x^{10} + 135x^9 - 540x^8 + 1214x^7 - 1482x^6 + 1161x^5 - 3240x^4 + 12960x^3 $$\displaystyle \ - 29136x^2 + 34992x - 17496 = 1000$
$\displaystyle x^{11}- 18x^{10} + 135x^9 - 540x^8 + 1214x^7 - 1482x^6 + 1161x^5 - 3240x^4 + 12960x^3 $$\displaystyle \ -29136x^2 + 34992x - 18496 = 0$
And that also yields 4 as it's only zero.