Question 673981
{{{4^(x+1)=12}}}..........since {{{4^(x+1)=4^x*4}}} replace {{{4^(x+1)}}} with {{{4^x*4}}}

{{{4^x*4=12}}}

{{{4^x=3}}}...now use a logarithm

{{{log(a,(x))= N}}}... means that in your case {{{a^N = x}}}.....{{{a=4}}}, {{{N=x}}} ,and {{{x=3}}}; so, we have

{{{log(4,( 3 ))= x}}}.......transfer into log base {{{10}}}


{{{log(3)/log(4)= x}}}

since {{{log(3 ) = 1.0986122886681098}}} and {{{log(4)= 1.3862943611198906188344642429163531361510002687205105}}}, we have

{{{x=1.0986122886681098/1.3862943611198906188344642429163531361510002687205105=0.7924812503605781690686400288008583335496142251280518}}}

{{{x=0.7924812503605781690686400288008583335496142251280518}}}...round it to two decimals

{{{x=0.79}}}