document.write( "Question 1209181: If 2^x = 12 and 6^y = 12, then what is (1/x) + (1/y) equal to? \n" ); document.write( "
Algebra.Com's Answer #847769 by math_tutor2020(3816)\"\" \"About 
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\n" ); document.write( "Answer: 1\r
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\n" ); document.write( "\n" ); document.write( "Explanation\r
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\n" ); document.write( "\n" ); document.write( "Let's isolate x so we can determine 1/x.
\n" ); document.write( "\"2%5Ex+=+12\"\r
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\n" ); document.write( "\n" ); document.write( "\"log%28%282%5Ex%29%29+=+log%28%2812%29%29\" Apply logs to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"x%2Alog%28%282%29%29+=+log%28%2812%29%29\" Use the rule log(M^N) = N*log(M) to pull down the exponent.\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+log%28%2812%29%29%2Flog%28%282%29%29\" Divide both sides by log(2).\r
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\n" ); document.write( "\n" ); document.write( "\"1%2Fx+=+log%28%282%29%29%2Flog%28%2812%29%29\" Apply the reciprocal to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"6%5Ey+=+12\" can be rearranged into \"1%2Fy+=+log%28%286%29%29%2Flog%28%2812%29%29\" when following similar steps.\r
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\n" ); document.write( "\n" ); document.write( "Then,
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\n" ); document.write( "\n" ); document.write( "\"1%2Fx+%2B+1%2Fy+=+%28+log%28%282%29%29++%2B+log%28%286%29%29+%29%2Flog%28%2812%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"1%2Fx+%2B+1%2Fy+=+%28+log%28%282%2A6%29%29+%29%2Flog%28%2812%29%29\" Use the rule log(M)+log(N) = log(M*N) where the base of each log must be the same. \r
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\n" ); document.write( "\n" ); document.write( "\"1%2Fx+%2B+1%2Fy+=+log%28%2812%29%29%2Flog%28%2812%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"1%2Fx+%2B+1%2Fy+=+1\"\r
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\n" ); document.write( "Another approach.\r
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\n" ); document.write( "\n" ); document.write( "2^x = 12 rearranges to 2 = 12^(1/x) when raising both sides to the 1/x power.
\n" ); document.write( "Similarly we can say that 6^y = 12 turns into 6 = 12^(1/y)\r
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\n" ); document.write( "\n" ); document.write( "We have these new equations
\n" ); document.write( "2 = 12^(1/x)
\n" ); document.write( "6 = 12^(1/y)\r
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\n" ); document.write( "\n" ); document.write( "These two new equations involve exponents 1/x and 1/y.
\n" ); document.write( "If we multiply straight down then we'll be able to add these exponents due to the rule a^b*a^c = a^(b+c).\r
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\n" ); document.write( "\n" ); document.write( "The left hand sides multiply to 12 aka 12^1.
\n" ); document.write( "The right hand sides multiply to 12^( (1/x) + (1/y) )
\n" ); document.write( "We form the equation 12^1 = 12^( (1/x) + (1/y) )\r
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\n" ); document.write( "\n" ); document.write( "Comparing the two sides shows that the bases are both 12, so the exponents must be equal.\r
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\n" ); document.write( "\n" ); document.write( "Therefore (1/x) + (1/y) = 1\r
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\n" ); document.write( "\n" ); document.write( "Let's confirm the answer. \r
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\n" ); document.write( "\n" ); document.write( "Use a calculator and the change of base formula to determine these two approximations
\n" ); document.write( "x = log2(12) = log(12)/log(2) = 3.584963
\n" ); document.write( "y = log6(12) = log(12)/log(6) = 1.386853
\n" ); document.write( "Each approximate value is rounded to 6 decimal places.\r
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\n" ); document.write( "\n" ); document.write( "Then,
\n" ); document.write( "(1/x) + (1/y) = (1/3.584963) + (1/1.386853)
\n" ); document.write( "(1/x) + (1/y) = 0.999999860928
\n" ); document.write( "We don't land on 1 exactly but we get close enough.
\n" ); document.write( "The rounding error will depend how you rounded the approximations for x and y. \r
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\n" ); document.write( "\n" ); document.write( "If you were to use more decimal digits in the values of x and y, then you'll get closer to 1.
\n" ); document.write( "Let's say we rounded to 12 decimal places (rather than 6)
\n" ); document.write( "(1/x) + (1/y) = (1/3.584962500721) + (1/1.386852807235)
\n" ); document.write( "(1/x) + (1/y) = 0.9999999999997738
\n" ); document.write( "We get much closer to 1 this time.
\n" ); document.write( "There are 12 copies of \"9\" listed in that number shown above.
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