SOLUTION: Solve: a) {{{ log(2^x) = 2 ^(logx) }}} b) {{{ e^x + e ^(-x) = 6 }}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve: a) {{{ log(2^x) = 2 ^(logx) }}} b) {{{ e^x + e ^(-x) = 6 }}}      Log On


   

Question 120725: Solve:
a) +log%282%5Ex%29+=+2+%5E%28logx%29+
b) +e%5Ex+%2B+e+%5E%28-x%29+=+6+
Answer by bucky(2189) About Me  (Show Source):
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
going to presume that I don't need to explain the details of each rule of logs that are
used. And I'm going to presume the logs are to the base 10 ...
.
(a) Given
.
log%282%5Ex%29+=+2%5E%28log%28x%29%29
.
The exponent on the left side can be brought out as a multiplier to give:
.
x%2Alog%282%29+=+2%5E%28log%28x%29%29
.
Take the log of both sides:
.
log%28x%2Alog%282%29%29+=+log%282%5E%28log%28x%29%29%29
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Apply the exponential rule to the right side:
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log%28x%2Alog%282%29%29+=+log%28x%29%2Alog%282%29
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Split the left side into two logs using the multiplication rule:
.
log%28x%29%2Blog%28log%282%29%29=log%28x%29%2Alog%282%29
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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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log%28x%29=log%28x%29%2Alog%282%29-log%28log%282%29%29
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Get rid of the log%28x%29%2Alog%282%29 on the right side by subtracting it from both sides to get:
.
log%28x%29-log%28x%29%2Alog%282%29-log%28log%282%29%29
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On the left side factor out the common term log%28x%29 and you have:
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log%28x%29%281-log%282%29%29=-log%28log%282%29%29
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Solve for log%28x%29 by dividing both sides of this equation by 1-log%282%29 to get:
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log%28x%29+=+-log%28log%282%29%29%2F%281-log%282%29%29
.
Now it's calculator time ... on the right side calculate the numerator by entering 2 and
taking the base 10 logarithm twice in succession. You should get -0.521390227
as the answer, but don't forget the minus sign preceding the numerator ... so the numerator
is positive and is 0.521390227. The denominator is 1-log%282%29 and a calculator will
tell you that that computes to be 0.698970004. This reduces the equation to:
.
log%28x%29+=+0.521390227%2F0.698970004=+0.745940776
.
So we have that:
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log%28x%29=0.745940776
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To solve for x convert from this logarithmic form to the exponential form as follows:
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x+=+10%5E0.745940776=+5.57109772
.
and that's the answer ... or as close to the correct answer as my calculator allows.
.
Note that x = 0 comes pretty close to being a correct answer also, but log%280%29 will
give you an error when you try to compute it. If you let x approach zero from the positive
side your given equation will become very close to equal on both sides. You can see this by
letting x be something such as 1 times 10^(-99) ... and plugging that very small value into
the given equation.
.
Next problem (b) Given:
.
+e%5Ex+%2B+e+%5E%28-x%29+=+6+
.
Multiply all terms on both sides by e%5Ex to get:
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+e%5E2x+%2B+e+%5E%280%29+=+6%2Ae%5Ex+
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but e%5E0+=+1 so the equation becomes:
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e%5E2x%2B1=6%2Ae%5Ex
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Subtract 6%2Ae%5Ex from both sides. The resulting equation is:
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e%5E2x+-6%2Ae%5Ex%2B1=0
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Just to simplify the problem a little, let's let y+=+e%5Ex. If we substitute y for e%5Ex
the equation becomes:
.
y%5E2%2B6y%2B1=0
.
This is an "ordinary" quadratic equation of standard form. Apply the quadratic formula to solve
it and you should get:
.

.
Using a calculator you should find that the two answers are:
.
y = 5.828427125 and y = 0.171572875
.
But recall that y=e%5Ex. So the two answers become:
.
e%5Ex+=+5.828427125 and
e%5Ex+=+0.171572875
.
Take the natural log of both sides. For the first answer you get:
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ln%28e%5Ex%29+=+ln%285.828427125%29
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The exponent on the left side becomes a multiplier and you have:
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x%2Aln%28e%29+=+ln%285.828427125%29
.
But ln%28e%29+=+1 so the equation is reduced to:
.
x+=+ln%285.828427125%29
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and a calculator will tell you that the right side is 1.762747174 so the answer is:
.
x+=+1.762747174
.
If you follow the exact same steps as above, but substitute 0.171572875 in place of 5.828427125,
you will get that the second answer for x (based on y equaling 0.171572875 this time) is:
.
x+=+ln%280.171572875%29+=+-1.762747174
.
Hope you can make sense of all this mess above and that you can learn something by diligently
working your way through the labyrinth ... Good luck