Question 471118
Another way of working this problem:
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{{{(log 16 + 2log 2)/log 4 = x}}}
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Convert 16 to 2^4 and 4 to 2^2
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{{{(log 2^4 + 2log 2)/log 2^2 = x}}}
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Bring the exponents out as multipliers of the logarithm terms
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{{{(4log2 +2log2)/2log 2 = x}}}
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Since both terms in the numerator involve log 2, the terms can be added together.
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{{{6 log2/2 log 2 = x}}}
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The log 2 in the numerator cancels with the log 2 in the denominator and you are left with:
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{{{6/2 = x}}}
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Which reduces to x = 3