# Log Prob.

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• Feb 3rd 2009, 04:20 PM
jimstewart_NC
Log Prob.
Hi,

I'm having difficulty beginning to solve the following log problem:

6^(2log6x + log6x) = 125
(Where the the two 6's are the base of the logs)

Any help is much appreciated.

This is what I have so far:

6^(log6x^2 + log6x) = 125

6^(log6 (x^2) (x) ) = 125

Am I on the right track here? Do I have to make the 6 and 125 the same base to solve?

Thanks,

Jim.
• Feb 3rd 2009, 04:37 PM
Soroban
Hello, Jim!

Quote:

$6^{2\log_6x + \log_6x} \:=\: 125$

This is what I have so far:

$6^{\log_6x^2 + \log_6x} \:=\:125$

$6^{\log_6(x^2) (x)} \:=\:125$

Am I on the right track here?

Then we have: . $\underbrace{6^{\log_6(x^3)}}_{\downarrow} \:=\:125$
. . . . . . . . . . . . . $x^3 \:=\:125$ .
. . . Do you see why?

. . . . . . . . . . . . . . $x \:=\:5$

• Feb 3rd 2009, 04:54 PM
jimstewart_NC
Hi Soroban,

Thank you for your reply.

I think I see why:

is it because 6^(log6) is equivalent to 1?

so 6^(log6(x^3)) turns into:

1 (x^3) (b/c
6^(log6) = 1)

and then just: x^3

and so you have:

x = 5 (b/c the cubed root of 125 is 5)

is this correct?
• Feb 3rd 2009, 05:14 PM
Mathnasium
Yep, looks good.