Question 744709
a)
{{{3^(2x-1) = 2^(x+2)}}}

Expanding exponents...

{{{3^(2x)*3^(-1) = 2^x*2^2}}}

simplifying...

{{{3^(2x)/3 = 2^x*4}}}

{{{(3^2)^x/3=4*2^x}}}

Next we isolate x...

{{{9^x/3=4*2^x}}}



{{{9^x/2^x=4*3}}}



Key step: we can combine the bases under the exponent since the exponent for both of them is the same...

{{{(9/2)^x=12}}}

{{{log((9/2)^x)=log12}}}

{{{x*log(9/2)=log12)}}}

{{{x=log12/log(9/2)}}}

which can be rewritten as...

{{{log12/(log9-log2)}}}


Less steps for the remaining parts...

b)
{{{ 2^(x+1) = 5^(1-x)}}}

{{{ 2^x*2 = 5*5^(-x)}}}

{{{2^x/5^x=5/2}}}

{{{(2/5)^x=5/2}}}

{{{(2/5)^x=5/2}}}

{{{(2/5)^x=(2/5)^(-1)}}}

after taking log of both sides and cancelling (i.e. if the bases are equal the the exponents are also equal to each other)



x=-1




Now with maximum efficiency we solve the last...

c)
{{{2^(x+1)/ 5^x = 3}}}

{{{(2*2^x)/5^x=3}}}

{{{(2/5)^x=3/2}}}


{{{log((2/5)^x)=log(3/2)}}}

{{{x=(log3-log2)/(log2-log5)}}}