# Thread: log9 (4) + log3(x) = 3 solve for x

1. ## log9 (4) + log3(x) = 3 solve for x

log9 (4) + log3(x) = 3 solve for x

by this I mean log base 9 (4) etc

The different bases makes this harder for me.

This is my (incorrect) attempt

log3(x) + log3(4) = 3
-------
log3(9)

log3(x) + log3(4/9) = 3

log3(4/9) = log(4/9) / log(3) = -0.35

log3(x) = 3 -- 0.35 = 3.35

3^3.35 = x = 39.66 which is incorrect.

Where did I go wrong and how do I solve this?

Angus

2. Originally Posted by angypangy
log9 (4) + log3(x) = 3 solve for x
Use these facts: $\displaystyle \log_9(4)=\frac{\ln(4)}{\ln(9)}~\&~\log_3(x)=\frac {\ln(x)}{\ln(3)}$

3. Originally Posted by angypangy
log9 (4) + log3(x) = 3 solve for x

by this I mean log base 9 (4) etc

The different bases makes this harder for me.

This is my (incorrect) attempt

log3(x) + log3(4) = 3
-------
log3(9)

log3(x) + log3(4/9) = 3

log3(4/9) = log(4/9) / log(3) = -0.35

log3(x) = 3 -- 0.35 = 3.35

3^3.35 = x = 39.66 which is incorrect.

Where did I go wrong and how do I solve this?

Angus
Your second equation confused me for a moment. It didn't print out the way you wanted it to. Here it is, just for clarity:
$\displaystyle log_3(x) + \frac{log_3(4)}{log_3(9)} = 3$

You applied your log rules incorrectly in the next line.
$\displaystyle \frac{log_3(4)}{log_3(9)} \neq log_3 \left ( \frac{4}{9} \right )$

But note that $\displaystyle log_3(9) = log_3(3^2)$. What can you do with this?

The rest of the method is correct, but unless you have any conflicting instructions, I'd solve it exactly...ie leave it in terms of log3(4). (Which you can actually simplify a bit.) There is no need to use a calculator on this one.

-Dan

4. Ah I looked at the log9 and thought I must be missing some trick with 3^2 = 9

OK so you just work out the logs. OK, no problem. Thanks.

5. Is this correct?

log3(4)
-------
log3(9)

= log3(4) - log3(9) = log3(4) - 2*log3(3) = log3(4) - 2

6. Originally Posted by angypangy
Is this correct?

log3(4)
-------
log3(9)

= log3(4) - log3(9) = log3(4) - 2*log3(3) = log3(4) - 2
This is correct, but I'd do it a bit differently. First of all, yes $\displaystyle log_3(9) = 2$. So....
$\displaystyle \frac{log_3(4)}{log_3(9)} = \frac{1}{2} log_3(4)$

The reason I would do so is that you can also simplify $\displaystyle log_3(4) = log_3(2^2)$

See what that gives you.

-Dan

7. Hello, angypangy!

$\text{Solve: }\;\log_94 + \log_3x \:=\:3$

I solved it like this . . .

$\text{Let: }\:\log_3x \,=\,P \quad\Rightarrow\quad 3^P \,=\,x \quad\Rightarrow\quad (9^{\frac{1}{2}})^P \,=\,x \quad\Rightarrow\quad 9^{\frac{P}{2}} \,=\,x$

$\text{Take logs, base 9: }\;\log_9(9^{\frac{P}{2}}) \,=\,\log_9x \quad\Rightarrow\quad \tfrac{P}{2}\log_99 \,=\,\log_9x$

. . $P \,=\,2\log_9x \quad\Rightarrow\quad \log_3x \,=\,2\log_9x$

$\text{The equation becomes: }\; \log_94 + 2\log_9x \:=\:3$

. . $\log_94 + \log_9x^2 \:=\:3 \quad\Rightarrow\quad \log_94x^2 \:=\:3$

. . $4x^2 \,=\,9^3 \quad\Rightarrow\quad x^2 \:=\:\tfrac{729}{4}$

$\text{Therefore: }\;x \:=\:\frac{27}{2}$

8. My final working out was like this:

log9(4) = log3(4)/log3(9) = log3(2^2)/log3(3^2) = 2*log3(2)/2*log3(3) - cancels down to log3(2)

=> log3(2) + log3(x) = 3

=> 3^3 = 2x => 2x = 27 => x = 27/2

9. Hi angypangy. I did this slightly differently The logB9(4)is converted to B3
log B9 (9) =1 Log B3 (9) =2 therefor log B9 (4) = 1/2 log B3 (4)

The final equations logB3 (2) + logB3(x) = log B3 (27)
log B3 (2x) = log B3 (27)
2x =27 x=27/2

bjh

10. Originally Posted by angypangy
My final working out was like this:

log9(4) = log3(4)/log3(9) = log3(2^2)/log3(3^2) = 2*log3(2)/2*log3(3) - cancels down to log3(2)

=> log3(2) + log3(x) = 3

=> 3^3 = 2x => 2x = 27 => x = 27/2

I would solve it similarly...

$\displaystyle log_94=y\Rightarrow\ 4=9^y\Rightarrow2^2=3^{2y}\Rightarrow\ 2=3^y$

$\displaystyle \Rightarrow\ log_32=y=log_94$

which is an alternative to changing base, due to the 9 being "3 squared".

Then

$\displaystyle log_32+log_3x=3=log_327$

$\displaystyle log_32x=log_327$

$\displaystyle 2x=27$

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