document.write( "Question 293954: Solve for x. Log2x – Log4x+ Log8x = 5 \n" ); document.write( "

Algebra.Com's Answer #212090 by stanbon(75887) $\"\"$ $\"About$

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Solve for x. Log2x – Log4x+ Log8x = 5
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\n" ); document.write( "Assuming that the 2,4, and 8 are base numbers you get:
\n" ); document.write( "Log2(x) - (1/2)Log2(x) + (1/3)log2(x) = 5
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\n" ); document.write( "log2(x) - log2(x^(1/2)) + log2(x^(1/3)) = 5
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\n" ); document.write( "log2[(x*x^(1/3))/x^(1/2)] = 5
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\n" ); document.write( "log2[x^(4/3-1/2)] = 5
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\n" ); document.write( "x^(5/8) = 5
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\n" ); document.write( "x = 5^(8/5)
\n" ); document.write( "x = 13.13..
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\n" ); document.write( "Assuming the 2x, 4x, and 8x are all inverse logs you get:
\n" ); document.write( "log(2x) - log(4x) + log(8x) = 5
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\n" ); document.write( "log[(2x)(8x)/(4x)] = 5
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\n" ); document.write( "log(4x) = 5
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\n" ); document.write( "4x = 10^5
\n" ); document.write( "x = (1/4)(100,000)
\n" ); document.write( "x = 250,000
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "

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