Question 160268
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{{{(125/8)^(x+1) = (2/5)^(x-1)}}}

Notice that {{{125=5^3}}} and that {{{8=2^3}}} so {{{125/8=(5/2)^3}}}
so replace {{{125/8}}} by {{{(5/2)^3}}}:

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

Multiply exponents on the left:

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

Distribute: 

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

Now observe that on the right the base {{{2/5}}} is the 
reciprocal of the base on the left, and the reciprocal
is the {{{-1}}} power, so write {{{2/5}}} on the right as {{{(5/2)^(-1)}}}:

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

Multiply the exponents on the right:

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

Distribute:

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

Since both sides are powers of the same base, and
since that base is neither 0 nor 1, we may equate
the exponents and drop the bases:

{{{3x+3=-x+1}}}

{{{4x=-2}}}

{{{x=-2/4}}}

{{{x=-1/2}}}

Edwin</pre>