# Thread: Problem with changing of base log question

1. ## Problem with changing of base log question

Hi all,
Just doing my revision and came across a logarithms question that involves changing of base. I could only complete halfway before getting stuck. So I hope someone here can help me out.

Heres the question:
3 log (base 3) x - 26= 9 log (base x) 3

I have managed to switch the base of 9 log from x to 3 but soon got stuck at the part whereby 9/log (base 3) x----doesn't matter if you don't understand, just solve it your own way.

2. $3 \log_3 {(x)} - 26= 9 \log_x {(3)}$

First isolate all the $x$ terms on a side :

$3 \log_3 {(x)} - 9 \log_x {(3)}= 26$

Note that you can factorize out a $3$ to make it simpler :

$3 (\log_3 {(x)} - 3 \log_x {(3)})= 26$

Divide to simplify :

$\log_3 {(x)} - 3 \log_x {(3)}= \frac{26}{3}$

Using the power log law :

$\log_3 {(x)} - \log_x {(9)}= \frac{26}{3}$

Now, $\log_x {(9)} = \frac{\log_3 {(9)}}{\log_3 {(x)}}$ (we're using 3 as a changing base for obvious reasons). Substituting :

$\log_3 {(x)} - \frac{\log_3 {(9)}}{\log_3 {(x)}}= \frac{26}{3}$

Now, $\log_3 {(9)} = 2$, so :

$\log_3 {(x)} - \frac{2}{\log_3 {(x)}}= \frac{26}{3}$

Setting $u = \log_3 {(x)}$, you only have to solve :

$u - \frac{2}{u} = \frac{26}{3}$

The solutions are given by substituting another well-chosen value for $u$ so as to find a quadratic equation. The solutions are :

$u = \frac{1}{3} \left ( 13 \pm \sqrt{187} \right )$

Now, since $u = \log_3 {(x)}$, $x = 3^u$. Just reverse the substitution to find the values of $x$, and you are done

Does it make sense ? (hope I'm right, Wolfram won't check it for some reason)

3. Originally Posted by AeroScizor
Hi all,
Just doing my revision and came across a logarithms question that involves changing of base. I could only complete halfway before getting stuck. So I hope someone here can help me out.

Heres the question:
3 log (base 3) x - 26= 9 log (base x) 3

I have managed to switch the base of 9 log from x to 3 but soon got stuck at the part whereby 9/log (base 3) x----doesn't matter if you don't understand, just solve it your own way.

Is it $3 \log_3 (x) - 26 = 9 \log_x 3$ or $3 \log_3 (x - 26) = 9 \log_x 3$?

4. To Bacterius:

5. Originally Posted by mr fantastic
Is it $3 \log_3 (x) - 26 = 9 \log_x 3$ or $3 \log_3 (x - 26) = 9 \log_x 3$?
It's the former, not the latter.

6. Originally Posted by AeroScizor
Hi all,
Just doing my revision and came across a logarithms question that involves changing of base. I could only complete halfway before getting stuck. So I hope someone here can help me out.

Heres the question:
3 log (base 3) x - 26= 9 log (base x) 3

I have managed to switch the base of 9 log from x to 3 but soon got stuck at the part whereby 9/log (base 3) x----doesn't matter if you don't understand, just solve it your own way.

hi AeroScizor,

a small correction is needed.

$3log_3x-26=9log_x3$

$3log_3x=log_x3^9+26$

$3log_3x-log_x3^9=26$

using the change of base formula $log_bx=\frac{log_ax}{log_ab}$ we get

$3log_3x-\frac{log_33^9}{log_3x}=26$

$log_33^9=9$

$\frac{3\left(log_3x\right)^2-9}{log_3x}=26$

$3\left(log_3x\right)^2-9=26log_3x$

$3a^2-26a-9=0$

$a=\frac{26\pm\sqrt{784}}{6}=\frac{26\pm28}{6}=\fra c{54}{6},\ -\frac{2}{6}$

$log_3x=9,\ -\frac{1}{3}$

There was a small typo in the earlier post, as $3log_x3=log_x3^3=log_x27$

7. Hello, AeroScizor!

Solve for $x\!:\;\;3\log_3(x) - 26 \;=\; 9\log_x(3)$
Formula: . $\log_b(a) \:=\:\frac{1}{\log_a(b)}$

So we have: . $3\log_3(x) - 26 \;=\;\frac{9}{\log_3(x)}$

Multiply by $\log_3(x)\!:\quad 3\bigg[\log_3(x)\bigg]^2 - \;26\log_3(x) \;=\;9 \quad\Rightarrow\quad3\bigg[\log_3(x)\bigg]^2 - \;26\log_3(x) \;-\; 9 \;=\;0$

Factor: . $\bigg[3\log_3(x) + 1\bigg]\,\bigg[\log_3(x) - 9\bigg] \;=\;0$

And we have two equations to solve:

. . $3\log_3(x) + 1 \:=\:0 \quad\Rightarrow\quad \log_3(x) \:=\:-\tfrac{1}{3} \quad \hdots\text{ No real roots}$

. . $\log_3(x) - 9 \:=\:0 \quad\Rightarrow\quad \log_3(x) \:=\:9 \quad\Rightarrow\quad x \:=\:3^9$

Therefore: . $x \;=\;19,\!683$