Hi;
Solve for x: 4log(x - 6) = 11
so far I'm hare log(x - 6)^4 = 11 (not sure where to go from here).
Thanks.
The inverse of $\displaystyle log_a(x)$ is $\displaystyle a^x$- that is if $\displaystyle y= log_a(x)$ then $\displaystyle x= a^y$. Your problem is a little ambiguous because you do not give a base for the logarithm. Normally, that would imply the "common logarithm", base 10. If that is the base here, then $\displaystyle (x- 6)^4= 10^{11}$. That is probably correct here but in higher level mathematics it is common to use "log" rather than "ln" to mean the natural logarithm. If that is the case, then $\displaystyle (x- 6)^4= e^{11}$.