Hi;
solve logbase6 36^5x.
6^2 = 36^5x
2 =5x therefore x=2/5.
Is that correct.I think it is
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:
$\displaystyle 10x=2$
$\displaystyle 5x=1$
$\displaystyle x=\frac{1}{5}$