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solve exactly for a log(base4)a^5+ log(base2)a^3/2 = log(base8) 625
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using the formula:\[\log_{b^n}(x)=\frac{1}{n}\log_b(x)\] you can rewrite all the logs in base 2.
or...now that im thinking about it....you could also write them in terms of log base 8 by using the formula backwards:\[\log_b(x)=nlog_{b^n}(x)\]
So using the fact that 2^3=8, and 4^(3/2)=8, you can say:\[\log_4(a^5)=\frac{3}{2}\log_8(a^5)\]and\[\log_2(a^{\frac{3}{2}})=3\log_8(a^\frac{3}{2})\]
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