Question 604786
given to solve for x:
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{{{6*4^x=99}}}
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First, take the base 10 log of both sides:
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{{{log(6*4^x) = log(99)}}}
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Using the product rule for logarithms, split the left side into two logarithms:
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{{{log(6) + log(4^x) = log(99)}}}
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Subtract log(6) from both sides to get:
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{{{log (4^x) = log(99) - log(6)}}}
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By the rules for logarithms, an exponent can be brought outside as a multiplier. Therefore, on the left side, bring the x out as a multiplier of the logarithm. The result is:
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{{{x*log(4) = log(99) - log(6)}}}
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Divide both sides by log(4):
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{{{x = (log(99) - log(6))/log(4)}}}
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Use a calculator to find that log(99) = 1.995635195, log(6) =  0.77815125, and log(4) = 0.602059991. Substitute these values into the equation for x to get:
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{{{x = (1.995635195 - 0.77815125)/0.602059991}}}
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Subtract the two terms on the numerator:
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{{{x = 1.217483945/0.602059991}}}
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Do the division on the right side and you have:
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{{{x = 2.022197062}}}
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The problem tells you to round this answer to two decimal places. When you do that, the answer shortens to:
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{{{x = 2.02}}}
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That's the answer that you're looking for. Hope this helps you to understand the process and the rules that apply to logarithms better.
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