Question

can anyine solve this logarithm question 3c

Answer 1

hello users .....

solution:-
we know that:

$log_{b} a = \frac{loga}{logb}$

And

$log m^{n} = n*log m$

Here,
we have;

$log_{8}[ 128 * 4^{3} ]$

= $\frac{log [128* 4^{3} ] }{log 8}$

=$\frac{log [ 2^{7} * 2^{6}] }{log [2^{3}] }$

=$\frac{log[ 2^{13}] }{log[2^{3}] }$

=$\frac{13*log 2}{3*log2}$ 

= 13 / 3 Answer 

# hope it helps :)