# How do you condense 1/2(log_4(x+1)+2log_4(x-1))+6log_4x?

Let $E = \frac{1}{2} \left[{\log}_{4} \left(x + 1\right) + 2 {\log}_{4} \left(x - 1\right)\right] + 6 {\log}_{4} x$ represent your expression (just so then I don't have to rewrite it so often!). Using the logarithm's identities:
$E = \frac{1}{2} \left[{\log}_{4} \left(x + 1\right) + {\log}_{4} {\left(x - 1\right)}^{2}\right] + {\log}_{4} {x}^{6}$;
$E = \frac{1}{2} {\log}_{4} \left[\left(x + 1\right) {\left(x - 1\right)}^{2}\right] + {\log}_{4} {x}^{6}$;
$E = {\log}_{4} \left[{\left(x + 1\right)}^{\frac{1}{2}} \left(x - 1\right)\right] + {\log}_{4} {x}^{6}$;
$E = {\log}_{4} \left[{\left(x + 1\right)}^{\frac{1}{2}} \left(x - 1\right) {x}^{6}\right]$.