document.write( "Question 120725: Solve: \r
\n" ); document.write( "\n" ); document.write( "a) \"+log%282%5Ex%29+=+2+%5E%28logx%29+\"
\n" ); document.write( "b) \"+e%5Ex+%2B+e+%5E%28-x%29+=+6+\"
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Algebra.Com's Answer #88532 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
These are a couple of complex problems. It might get a little tricky to follow the math. I'm
\n" ); document.write( "going to presume that I don't need to explain the details of each rule of logs that are
\n" ); document.write( "used. And I'm going to presume the logs are to the base 10 ...
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\n" ); document.write( "(a) Given
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\n" ); document.write( "\"log%282%5Ex%29+=+2%5E%28log%28x%29%29\"
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\n" ); document.write( "The exponent on the left side can be brought out as a multiplier to give:
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\n" ); document.write( "\"x%2Alog%282%29+=+2%5E%28log%28x%29%29\"
\n" ); document.write( ".
\n" ); document.write( "Take the log of both sides:
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\n" ); document.write( "\"log%28x%2Alog%282%29%29+=+log%282%5E%28log%28x%29%29%29\"
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\n" ); document.write( "Apply the exponential rule to the right side:
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\n" ); document.write( "\"log%28x%2Alog%282%29%29+=+log%28x%29%2Alog%282%29\"
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\n" ); document.write( "Split the left side into two logs using the multiplication rule:
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\n" ); document.write( "\"log%28x%29%2Blog%28log%282%29%29=log%28x%29%2Alog%282%29\"
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\n" ); document.write( "Get rid of \"log%28log%282%29%29\" on the left side by subtracting it from both sides to result in:
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\n" ); document.write( "\"log%28x%29=log%28x%29%2Alog%282%29-log%28log%282%29%29\"
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\n" ); document.write( "Get rid of the \"log%28x%29%2Alog%282%29\" on the right side by subtracting it from both sides to get:
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\n" ); document.write( "\"log%28x%29-log%28x%29%2Alog%282%29-log%28log%282%29%29\"
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\n" ); document.write( "On the left side factor out the common term \"log%28x%29\" and you have:
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\n" ); document.write( "\"log%28x%29%281-log%282%29%29=-log%28log%282%29%29\"
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\n" ); document.write( "Solve for \"log%28x%29\" by dividing both sides of this equation by \"1-log%282%29\" to get:
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\n" ); document.write( "\"log%28x%29+=+-log%28log%282%29%29%2F%281-log%282%29%29\"
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\n" ); document.write( "Now it's calculator time ... on the right side calculate the numerator by entering 2 and
\n" ); document.write( "taking the base 10 logarithm twice in succession. You should get \"-0.521390227\"
\n" ); document.write( "as the answer, but don't forget the minus sign preceding the numerator ... so the numerator
\n" ); document.write( "is positive and is \"0.521390227\". The denominator is \"1-log%282%29\" and a calculator will
\n" ); document.write( "tell you that that computes to be \"0.698970004\". This reduces the equation to:
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\n" ); document.write( "\"log%28x%29+=+0.521390227%2F0.698970004=+0.745940776\"
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\n" ); document.write( "So we have that:
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\n" ); document.write( "\"log%28x%29=0.745940776\"
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\n" ); document.write( "To solve for x convert from this logarithmic form to the exponential form as follows:
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\n" ); document.write( "\"x+=+10%5E0.745940776=+5.57109772\"
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\n" ); document.write( "and that's the answer ... or as close to the correct answer as my calculator allows.
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\n" ); document.write( "Note that x = 0 comes pretty close to being a correct answer also, but \"log%280%29\" will
\n" ); document.write( "give you an error when you try to compute it. If you let x approach zero from the positive
\n" ); document.write( "side your given equation will become very close to equal on both sides. You can see this by
\n" ); document.write( "letting x be something such as 1 times 10^(-99) ... and plugging that very small value into
\n" ); document.write( "the given equation.
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\n" ); document.write( "Next problem (b) Given:
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\n" ); document.write( "\"+e%5Ex+%2B+e+%5E%28-x%29+=+6+\"
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\n" ); document.write( "Multiply all terms on both sides by \"e%5Ex\" to get:
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\n" ); document.write( "\"+e%5E2x+%2B+e+%5E%280%29+=+6%2Ae%5Ex+\"
\n" ); document.write( ".
\n" ); document.write( "but \"e%5E0+=+1\" so the equation becomes:
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\n" ); document.write( "\"e%5E2x%2B1=6%2Ae%5Ex\"
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\n" ); document.write( "Subtract \"6%2Ae%5Ex\" from both sides. The resulting equation is:
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\n" ); document.write( "\"e%5E2x+-6%2Ae%5Ex%2B1=0\"
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\n" ); document.write( "Just to simplify the problem a little, let's let \"y+=+e%5Ex\". If we substitute \"y\" for \"e%5Ex\"
\n" ); document.write( "the equation becomes:
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\n" ); document.write( "\"y%5E2%2B6y%2B1=0\"
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\n" ); document.write( "This is an \"ordinary\" quadratic equation of standard form. Apply the quadratic formula to solve
\n" ); document.write( "it and you should get:
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\n" ); document.write( "Using a calculator you should find that the two answers are:
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\n" ); document.write( "y = 5.828427125 and y = 0.171572875
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\n" ); document.write( "But recall that \"y=e%5Ex\". So the two answers become:
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\n" ); document.write( "\"e%5Ex+=+5.828427125\" and
\n" ); document.write( "\"e%5Ex+=+0.171572875\"
\n" ); document.write( ".
\n" ); document.write( "Take the natural log of both sides. For the first answer you get:
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\n" ); document.write( "\"ln%28e%5Ex%29+=+ln%285.828427125%29\"
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\n" ); document.write( "The exponent on the left side becomes a multiplier and you have:
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\n" ); document.write( "\"x%2Aln%28e%29+=+ln%285.828427125%29\"
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\n" ); document.write( "But \"ln%28e%29+=+1\" so the equation is reduced to:
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\n" ); document.write( "\"x+=+ln%285.828427125%29\"
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\n" ); document.write( "and a calculator will tell you that the right side is 1.762747174 so the answer is:
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\n" ); document.write( "\"x+=+1.762747174\"
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\n" ); document.write( "If you follow the exact same steps as above, but substitute 0.171572875 in place of 5.828427125,
\n" ); document.write( "you will get that the second answer for x (based on y equaling 0.171572875 this time) is:
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\n" ); document.write( "\"x+=+ln%280.171572875%29+=+-1.762747174\"
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\n" ); document.write( "Hope you can make sense of all this mess above and that you can learn something by diligently
\n" ); document.write( "working your way through the labyrinth ... Good luck
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