# Log Prob.

• Feb 3rd 2009, 03:20 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, 03:37 PM
Soroban
Hello, Jim!

Quote:

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

This is what I have so far:

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

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

Am I on the right track here?

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

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

• Feb 3rd 2009, 03:54 PM
jimstewart_NC
Hi Soroban,

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, 04:14 PM
Mathnasium
Yep, looks good.