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625^x=3125
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\[x=\frac{\ln(3125)}{\ln(625)}\]
you solve \[b^x=A\] for x in one step via \[x=\frac{\ln(A)}{\ln(b)}\]
if 3125 is an integer power of x then you can guess perhaps
i mean if 3125 is an integer power of 625
but it is not
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ok so now what?
625 = 5^4, 3125 = 5^5. Take log(5) of both sides to get x log(5)(5^4) = log(5)(5^4) 4x = 4 x = 1.
oops x log(5)(5^4) = log(5)(5^5) 4x = 5 x = 5/4
too much work. whip out mr calculator and type in \[\frac{\ln(3125)}{\ln(625)}\] you will get \[\frac{5}{4}\]
although of course abtrhearn is right. but if you recognize that \[625=5^4\] and \[3125=5^5\] you do not need logs. just write \[5^{4x}=5^5\] so \[4x=5\] and \[x=\frac{5}{4}\]
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