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Math Help - Math book error? Exponential Equation..

  1. #1
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    Math book error? Exponential Equation..

    4(2^x)+31(2^x)-8=0

    OK, I have a good feeling it's going to turn into a complex trinomial.
    By substituting (2^x) with a variable, say "z"

    That would give us..

    4z+31z-8=0

    Still not a trinomial..hmm
    I see that the factor may be, -1 and 32.

    Or, do I somehow break down the 4 into 2^2. But still the exponent 2 would not be over the z.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NotSoBasic View Post
    4(2^x)+31(2^x)-8=0

    OK, I have a good feeling it's going to turn into a complex trinomial.
    By substituting (2^x) with a variable, say "z"

    That would give us..

    4z+31z-8=0

    Still not a trinomial..hmm
    I see that the factor may be, -1 and 32.

    Or, do I somehow break down the 4 into 2^2. But still the exponent 2 would not be over the z.
    Well, 4(2^x)+31(2^x)-8=0\implies 35(2^x)=8\implies 2^x=\frac{8}{35}.

    Now apply logs to both sides and solve for x. Can you continue?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Well, 4(2^x)+31(2^x)-8=0\implies 35(2^x)=8\implies 2^x=\frac{8}{35}.

    Now apply logs to both sides and solve for x. Can you continue?
    I'll try. Don't think I've had to do this before..

    log2^x=log\frac{8}{35}

    That fraction seems like it cannot be reduced..Unless..

    log2^x=log8-log35

    log2^x=log2^3-log35

    Hmm..??
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NotSoBasic View Post
    I'll try. Don't think I've had to do this before..

    log2^x=log\frac{8}{35}

    That fraction seems like it cannot be reduced..Unless..

    log2^x=log8-log35

    log2^x=log2^3-log35

    Hmm..??
    You don't need to reduce the fraction.

    However, the key to this is to note that \log 2^{\color{red}x}={\color{red}x}\log 2.

    Then your equation becomes x\log 2=\log\frac{8}{35}\implies x=\dots

    Can you finish off?
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    You don't need to reduce the fraction.

    However, the key to this is to note that \log 2^{\color{red}x}={\color{red}x}\log 2.

    Then your equation becomes x\log 2=\log\frac{8}{35}\implies x=\dots

    Can you finish off?
    Oh OK. That is the Law of Logarithms for Powers (LLP)
    logap^n = nlogap
    I've used that before. =]

    That fraction is REALLY throwing me off..
    x\log 2=\log\frac{8}{35}

    Next, I think you divide the both sides by log2?
    Or do I get the values, like:

    x(.3010)=(-.6409)
    x=-2.129

    It sorta works, and when subbed into the original equation it equals 0.00156

    There must be a better way..
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NotSoBasic View Post
    Oh OK. That is the Law of Logarithms for Powers (LLP)
    logap^n = nlogap
    I've used that before. =]

    That fraction is REALLY throwing me off..
    x\log 2=\log\frac{8}{35}

    Next, I think you divide the both sides by log2?
    Or do I get the values, like:

    x(.3010)=(-.6409)
    x=-2.129

    It sorta works, and when subbed into the original equation it equals 0.00156

    There must be a better way..
    Don't let the fraction get to you!

    When you take the log of a fraction, the value is negative.

    So it is ok to right the solution as x=\frac{\log\frac{8}{35}}{\log 2}. Now by change of base formula \log_ab=\frac{\log b}{\log a}, we can rewrite our solution as x=\log_2\left(\frac{8}{35}\right).

    Now when you plug it into verify the solution, we keep in mind that a^{\log_ab}=b:

    4\left(2^{\log_2\left(\frac{8}{35}\right)}\right)+  31\left(2^{\log_2\left(\frac{8}{35}\right)}\right)-8=4\left(\frac{8}{35}\right)+31\left(\frac{8}{35}\  right)-8=35\left(\frac{8}{35}\right)-8=8-8=0.

    So it shows that x=\frac{\log\frac{8}{35}}{\log 2}=\log_2\left(\frac{8}{35}\right) is the solution to our equation.
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  7. #7
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    Quote Originally Posted by NotSoBasic View Post
    4(2^x)+31(2^x)-8=0

    OK, I have a good feeling it's going to turn into a complex trinomial.
    By substituting (2^x) with a variable, say "z"

    That would give us..

    4z+31z-8=0

    Still not a trinomial..hmm
    The problem as given is probably correct, but since you mention trinomials, I have also seen problems where one of the exponents is 2x instead of x, like this:
    2^{{\color{red}2x}} + 6\cdot 2^x + 8 = 0.
    So, if you substitute (2^x) with another variable (I'll use y), you would get
    y^2 + 6y + 8 = 0.

    I'd double check wherever you got the problem to make sure that the exponents are correct.


    01
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    Don't let the fraction get to you!

    When you take the log of a fraction, the value is negative.

    So it is ok to right the solution as x=\frac{\log\frac{8}{35}}{\log 2}. Now by change of base formula \log_ab=\frac{\log b}{\log a}, we can rewrite our solution as x=\log_2\left(\frac{8}{35}\right).

    Now when you plug it into verify the solution, we keep in mind that a^{\log_ab}=b:

    4\left(2^{\log_2\left(\frac{8}{35}\right)}\right)+  31\left(2^{\log_2\left(\frac{8}{35}\right)}\right)-8=4\left(\frac{8}{35}\right)+31\left(\frac{8}{35}\  right)-8=35\left(\frac{8}{35}\right)-8=8-8=0.

    So it shows that x=\frac{\log\frac{8}{35}}{\log 2}=\log_2\left(\frac{8}{35}\right) is the solution to our equation.
    That's intense..Good job..

    It does appear to work out.
    I don't recall doing an equation of this complexity before. Maybe it's one of those "test" questions to make you really scratch your head.

    Thanks, I've learned alot.
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