Question 1209354
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{{{log(8a,(4b))=(log((4b)))/(log((8a)))}}} Change of base rule.


{{{log(8a,(4b))=(log((4*32)))/(log((8*1)))}}} Plug in a = 1 and b = 32.


{{{log(8a,(4b))=(log((128)))/(log((8)))}}}


{{{log(8a,(4b))=(log((2^7)))/(log((2^3)))}}} Rewrite 128 and 8 as powers of 2.


{{{log(8a,(4b))=(7*log((2)))/(3*log((2)))}}} Use the rule log(A^B) = B*log(A)


{{{log(8a,(4b))=7/3}}}


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Another approach


{{{log(8a,(4b))=c}}}


{{{4b = (8a)^c}}}


{{{4*32 = (8*1)^c}}}


{{{128 = 8^c}}}


{{{2^7 = (2^3)^c}}}


{{{2^7 = 2^(3c)}}}


{{{7 = 3c}}} Since the bases of the previous equation are equal, the exponents must be equal.


{{{c = 7/3}}}


{{{log(8a,(4b))=7/3}}}


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Answer: <font color=red>7/3</font>
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