Question 1105705
if  {{{ (b/a)^n = a }}}

and  {{{ (b/a)^x = b }}}

then  {{{ highlight(x = n+1) }}}

————
Check:  

{{{ (b/a)^(n+1) = (b/a)^n * (b/a) = a(b/a) = b }}}   

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EDIT 1/2/18, to the student:
I will get you started, and you should strive to figure out how I got to the solution on your own.
There may be a more efficient way to solve… this is just how I did it:

{{{ (b/a)^n = a }}}  —>   {{{ b = a^(1+1/n) }}}   (by taking nth root of both sides, then mult by 'a' on both sides)
Now starting with
{{{ (b/a)^x = b }}}
Substitute for 'b'
{{{ (a^(1+1/n) / a)^x = a^(1+1/n) }}}
{{{ (a^(1/n))^x = a^(1+1/n) }}}
Take log() of both sides:
 {{{ x * log((a^(1/n))) =  (1+1/n)log((a)) }}} 
 {{{ x * (1/n)cross(log((a))) = (1+1/n)*cross(log((a)) )}}}

so x = n + 1