document.write( "Question 61154: I need helping solving this for x:
\n" ); document.write( "(ln x)^3=ln x^4
\n" ); document.write( "Can I rewrite it as (ln x)^3=4lnx ?
\n" ); document.write( "Then can I divide both sides by lnx leaving (ln x)^2=4 ?
\n" ); document.write( "Can I now square both sides leaving me with ln x=2 ?
\n" ); document.write( "Now I'm not sure what to do next.
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Algebra.Com's Answer #42160 by mathick(4)\"\" \"About 
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I need helping solving this for x:
\n" ); document.write( "(ln x)^3=ln x^4
\n" ); document.write( "Can I rewrite it as (ln x)^3=4lnx ?\r
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\n" ); document.write( "\n" ); document.write( "Yes.\r
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\n" ); document.write( "\n" ); document.write( "Then can I divide both sides by lnx leaving (ln x)^2=4 ?\r
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\n" ); document.write( "\n" ); document.write( "Yes, but this assumes that you're not dividing both sides by 0, i.e. that ln(x) is not 0. This step wouldn't be valid in the case that ln(x) = 0, so this case (ln(x) = 0) needs to be treated separately.\r
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\n" ); document.write( "\n" ); document.write( "Can I now square (root) both sides leaving me with ln x=2 ?\r
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\n" ); document.write( "\n" ); document.write( "Right, ln (x) = 2, and also ln(x) = -2. (Taking the square root of both sides gives ln x = +2 and ln x = -2.) \r
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\n" ); document.write( "\n" ); document.write( "Now I'm not sure what to do next\r
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\n" ); document.write( "\n" ); document.write( "To solve ln(x) = 2 for x, exponentiate both sides:\r
\n" ); document.write( "\n" ); document.write( "\"e%5E%28ln%28x%29%29+=+e%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "The left side simplifies, giving one of the final answers:\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+e%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "The equation ln (x) = -2 can be solved similarly.\r
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\n" ); document.write( "\n" ); document.write( "Finally, there is the case when ln(x) = 0. This happens when x = 1. To verify that this is a solution, you can plug it into the original equation and see if it checks out (gives a true equation).
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