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Math Help - exponential equation

  1. #1
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    exponential equation

    32 x (8*) = 3 ; give an exact solution for * (* is a power)


    any help would be appreciated
    Last edited by highsamurai; November 20th 2008 at 07:52 PM.
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  2. #2
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    Quote Originally Posted by highsamurai View Post
    32 x (8*) = 3 ; give an exact solution for *


    any help would be appreciated
    I don't know what class this is for, but this is a good basic algebra problem. I would stop using "x" to mean multiply because in algebra you'll use "x" to mean an unknown number most of the time. You can use the "*" as a symbol for multiplication, use parentheses like 8(4) to mean 8 times 4.

    Let "x" be the number that we want to find that fits the problems guidelines. It doesn't matter what it is yet, but we know it is a real number, so just call it x.

    So 32*8*x=3 meaning 32 times 8 times x equals 3. We want to rearrange this to get this in the form of x=... so let's do that. The opposite operation of multiplication is division, so if we want to "cancel" the 32, we can divide by 32 on both sides of the equation and still keep the = sign. Do the same to the 8 and you should get x=\frac{3}{32*8}
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  3. #3
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    i guess i should have been more specific about the problem. the "*" is a root actually not an "x".

    32 x (8*) = 3 or 8* = 3/32


    x = multiply

    *= power

    thnx though
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  4. #4
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    Hello, highsamurai!

    Is that an exponent?


    Solve for x\!:\quad32\cdot8^x \:=\: 3

    We have: . 2^5(2^3)^x \:=\:3 \quad\Rightarrow\quad 2^{3x+5} \:=\:3

    Take logs: . \ln\left(2^{3x+5}\right) \:=\:\ln(3) \quad\Rightarrow\quad(3x+5)\ln(2) \:=\:\ln(3)

    . . 3x+5 \:=\:\frac{\ln(3)}{\ln(2)} \quad\Rightarrow\quad 3x \:=\:\frac{\ln(3)}{\ln(2)} - 5 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{3}\left(\frac{\ln(3)}{\ln(2)} - 5\right)}

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  5. #5
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    thnx man that helped a lot.


    I also have another one if you dont mind. I just dont know what to do next.

    log10 (x - 5) + log10 (3x + 6) = log10 (x+4)

    log10 (x - 5)(3x + 6) = log10 (x + 4)

    (x - 5)(3x + 6) = (x + 4)

    3x^2 - 9x - 30 = x + 4

    3x^2 - 9x - 34 = x

    (the ^2 is squared)
    Im lost after that, i dont know how to factor that.

    (the "10's" should be lowered)
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