Logarithmic equation

**james_bond** · March 17th 2008, 07:18 AM

Solve for $x$ :

$\log_{2008}\left(x-3\right)+\log_{2009}\left(x-3\right)=3-\lg\left(x^5-24\right)$

**Aryth** · March 17th 2008, 02:54 PM

Well, we are given:

$log_{2008}{(x - 3)} + log_{2009}{(x - 3)} = 3 - log(x^5 - 24)$

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:

$log_a{b} = \frac{log{b}}{log{a}}$

Where we are using base 10 because the right side has a base 10 logarithm.

Therefore we get:

$\frac{1}{log{2008}}log{(x - 3)} + \frac{1}{log{2009}}log{(x - 3)} = 3 - log{(x^5 - 24)}$

Where:

$log{2008} = 3.203$

and

$log{2009} = 3.203$

(They are very close together, where the first difference is in the ten thousandths place.)

So we get:

$0.31223log{(x - 3)} + 0.31223log{(x - 3)} = 0.31223(log{(x - 3)^2})$

$0.62446log{(x - 3)} = 3 - log{(x^5 - 24)}$

There's really not an easy way to solve this, so I'm going to make the following approximation:

$0.62446 \approx \frac{2}{3}$

$log{(x - 3)^{\frac{2}{3}}} = 3 - log(x^5 - 24)$

$log{(x - 3)^{\frac{2}{3}}} + log(x^5 - 24) = 3$

$log{(x - 3)^{\frac{2}{3}}(x^5 - 24)} = 3$

$(x - 3)^{\frac{2}{3}}(x^5 - 24) = 10^3$

$\sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2}$

$(x^5 - 24) = \frac{1000}{(x - 3)^5}$

$(x^5 - 24)(x + (-3))^5 = 1000$

$(x^5 - 24)(x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243) = 1000$

$x^{10} - 15x^9 + 90x^8 - 270x^7 + 405x^6 - 267x^5 + 360x^4 - 2160^3 + 6480x^2 - 9720x + 4832 = 0$

You get:

$x = 4$

To verify:

$ \frac{\log{1}}{log{2008}}+ \frac{\log{1}}{log{2009}} = 3 - \log\left(1000\right) $

$0 + 0 = 3 - 3 = 0$

And there you go.

**topher0805** · March 17th 2008, 11:16 PM

$ \sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2} $

How did you get from that to the step below? It just doesn't make any sense...

$ (x^5 - 24) = \frac{1000}{(x - 3)^5} $

**Aryth** · March 18th 2008, 06:49 AM

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:

$ \sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2} $

$(\sqrt[3]{x^5 - 24})^3 = \left(\frac{10}{(x - 3)^2}\right)^3$

Evaluated, it equals:

$(x^5 - 24) = \frac{10^3}{((x - 3)^2)^3} = \frac{1000}{(x - 3)^6}$

But nonetheless, 4 is still the answer.

**topher0805** · March 18th 2008, 07:32 AM

It's interesting that you get the correct answer despite making an error...

**james_bond** · March 18th 2008, 09:14 AM

Just to clarify: $x=4$ is a solution. There can't be any other solutions because the LHS is strictly increasing but RHS is strictly decreasing because the fuction $\log_bx$ is stricly increasing if $b>1$ ( $x>0$ ).

**Aryth** · March 18th 2008, 09:17 AM

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:

$(x + (-3))^6$

$= x^6 - 18x^5 + 135x^4 - 540x^3 + 1214x^2 - 1458x + 729$

Now if we take $(x^5 - 24)$ and multiply we get:

$x^{11} - 18x^{10} + 135x^9 - 540x^8 + 1214x^7 - 1482x^6 + 1161x^5 - 3240x^4 + 12960x^3$ $\ - 29136x^2 + 34992x - 17496 = 1000$

$x^{11}- 18x^{10} + 135x^9 - 540x^8 + 1214x^7 - 1482x^6 + 1161x^5 - 3240x^4 + 12960x^3$ $\ -29136x^2 + 34992x - 18496 = 0$

And that also yields 4 as it's only zero.

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