Hi;

solve logbase6 36^5x.

6^2 = 36^5x

2 =5x therefore x=2/5.

Is that correct.I think it is

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- Sep 27th 2012, 10:18 AManthonyelogarithm problem
Hi;

solve logbase6 36^5x.

6^2 = 36^5x

2 =5x therefore x=2/5.

Is that correct.I think it is - Sep 27th 2012, 10:23 AManthonyeRe: logarithm problem
Or do both sides need to have same base therefore

6^2 = (6^2)^5x

giving 6^2 = 6^10x

resulting in 2 = 10x therefore x = 2/10

Yeah that looks better. - Sep 27th 2012, 12:00 PMMarkFLRe: logarithm problem
Presumably, the equation is:

$\displaystyle \log_6(36^{5x})=2$

Converting from logarithmic to exponential form is one way we could proceed, but here is another:

We may use the log property $\displaystyle \log_a(b^c)=c\cdot\log_a(b)$ to write:

$\displaystyle 5x\cdot\log_6(36)=2$

$\displaystyle 5x\cdot\log_6(6^2)=2$

Using the above property again, we have:

$\displaystyle 10x\cdot\log_6(6)=2$

Since $\displaystyle \log_a(a)=1$ we now have:

$\displaystyle 10x=2$

$\displaystyle 5x=1$

$\displaystyle x=\frac{1}{5}$ - Sep 27th 2012, 12:03 PManthonyeRe: logarithm problem
Great thank you just needed to make sure.

- Sep 27th 2012, 12:09 PMMarkFLRe: logarithm problem
Another way:

$\displaystyle \log_6(36^{5x})=2=2\cdot\log_6(6)=\log_6(6^2)=\log _6(36^1)$

Hence:

$\displaystyle 5x=1$

$\displaystyle x=\frac{1}{5}$ - Sep 27th 2012, 03:37 PMbjhopperRe: logarithm problem
another way

6^2= (6^2)^5x

36 = 36^5x

5x=1