SOLUTION: Solve the equation algebraically. {{{2^x+2^(-x)=5}}} I've tried almost everything I can think of, and still no luck! Please help!

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve the equation algebraically. {{{2^x+2^(-x)=5}}} I've tried almost everything I can think of, and still no luck! Please help!      Log On


   

Question 90951This question is from textbook
: Solve the equation algebraically.
2%5Ex%2B2%5E%28-x%29=5
I've tried almost everything I can think of, and still no luck! Please help! This question is from textbook

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Tricky problem:
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2%5Ex%2B2%5E%28-x%29=5
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Let's convert the term with the minus exponent to fractional form:
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2%5Ex+%2B+1%2F%282%5Ex%29+=+5
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Next, get rid of the denominator by multiplying both sides (all terms) by 2%5Ex. When
you do that multiplication the equation becomes:
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%282%5Ex%2A2%5Ex%29+%2B+2%5Ex%281%2F2%5Ex%29+=+5%2A2%5Ex
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which becomes:
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2%5E%28x%2Bx%29+%2B+cross%282%5Ex%29%2A1%2Fcross%282%5Ex%29+=+5%2A2%5Ex
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which simplifies to:
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2%5E%282x%29%29+%2B+1+=+5%2A2%5Ex
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Now recognize that 2%5E%282x%29 is equal to %282%5Ex%29%5E2+ by the power rule of exponents.
Substituting this makes the equation become:
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%282%5Ex%29%5E2+%2B+1+=+5%2A2%5Ex
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Then subtracting 5%2A%282%5Ex%29+ from both sides results in:
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%282%5Ex%29%5E2+-+5%2A%282%5Ex%29+%2B+1+=+0
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Now let 2%5Ex+=+t ... just because it helps to clarify what's coming next. Substituting
t for 2%5Ex gives:
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t%5E2+-+5t+%2B+1+=+0
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Notice that this is just an "ordinary" quadratic equation. If you apply the quadratic
formula to this problem you get two answers for t as follows:
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t+=+-%28-5%29%2F%282%2A1%29+%2B-+sqrt%28%28-5%29%5E2+-4%2A1%2A1%29%2F%282%2A1%29
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This simplifies to:
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t+=+5%2F2+%2B-+sqrt%2825-4%29%2F2+=+5%2F2+%2B-+sqrt%2821%29%2F2
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So we have two possible answers for t:
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t+=+5%2F2+%2B+sqrt%2821%29%2F2
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and
t+=+5%2F2+-+sqrt%2821%29%2F2
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Let's make life a little easier by converting these answers to decimals by using a calculator.
If you do the work you should get:
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t+=+5%2F2+%2B+sqrt%2821%29%2F2+=+2.5+%2B+4.582575695%2F2=2.5+%2B+2.291287847+=+4.791287847
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and
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t+=+5%2F2+-+sqrt%2821%29%2F2+=+2.5+-+4.582575695%2F2=2.5+-+2.291287847+=+0.208712152
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There they are ... two possible answers for t. But let's recall that t is equal to 2%5Ex.
So we can write our two answers as:
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2%5Ex+=+4.791287847 and 2%5Ex+=+0.208712152
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Now apply logs to solve for x. We could use any logs of any base, but for grins let's use
the natural logs ... indicated by ln .... on both equations:
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Start with the equation 2%5Ex+=+4.791287847 and take the ln of both sides to get:
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ln%282%5Ex%29+=+ln%284.791287847%29
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on the left side we can make the exponent x a multiplier to convert the equation to:
x%2Aln%282%29+=+ln%284.791287847%29
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Now use a calculator to determine that ln(2) = 0.69314718 and ln(4.791287847) = 1.566799237.
Substituting these values for the logarithms results in the equation becoming:
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x%2A%280.69314718%29+=+1.566799237
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Divide both sides by 0.6931471 and you get:
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x+=+1.566799237%2F0.6931471+=+2.260413752
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One possible value for x
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The other value for x is basically worked the same way ... just different numbers. Start
with 2%5Ex+=+0.208712152.
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Take natural logs of both sides:
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ln%282%5Ex%29+=+ln%280.208712152%29
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Bring the exponent out:
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x%2Aln%282%29=+ln%280.208712152%29
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Use a calculator to find the two natural logs:
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x%2A%280.69314718%29=+-1.566799239
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Divide both sides by 0.69314718
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x+=+-1.566799239%2F0.69314718+=+-2.260413495
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Check every step of the math above just to make sure I didn't read the calculator
wrong somewhere. I only know that x = 2.260413 and x = -2.260413 are pretty close to correct.
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Hope this helps you plow through the problem OK. Time for a coffee break.
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