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Math Help - exponetial equations

  1. #1
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    exponetial equations

    Solve in exact form

    1) 3(2)^x = 18^(x-1)

    2) 2(27)^x = 9^(x+1)

    Can anyone help with any of these questions?
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by imthatgirl View Post
    Solve in exact form

    1) 3(2)^x = 18^(x-1)

    2) 2(27)^x = 9^(x+1)

    Can anyone help with any of these questions?
    3(2^{x}) = 18^{x-1}

    Take the natural log of both sides:
    ln(3(2^{x})) = ln(18^{x-1})

    Use the multiplication identity: ln(ab) = ln(a) + ln(b)
    ln(3)+ ln(2^{x}) = ln(18^{x-1})

    Use the exponent identity: ln(a^{b}) = b\cdot ln(a)
    ln(3)+ x\cdot ln(2) = (x-1)ln(18)

    Distribute the ln(18)
    ln(3)+ x\cdot ln(2) = x\cdot ln(18) - ln(18)

    Put all the terms with x in them on one side by subtracting x*ln(2) from both sides:
    ln(3) = x\cdot ln(18) - ln(18) - x\cdot ln(2)

    Put all the terms without x in them on the other side by adding ln(18) to both sides:
    ln(3) + ln(18) = x\cdot ln(18) - x\cdot ln(2)

    Factor out an x:
    ln(3) + ln(18) = x[ln(18) - ln(2)]

    Divide by the coefficient of x, which is [ln(18)-ln(2)]
    \frac{ln(3) + ln(18)}{ln(18) - ln(2)} = x

    Use additive identity of logarithms to see that ln(3)+ln(18)=ln(3\cdot 18)=ln(54)
    \frac{ln(54)}{ln(18) - ln(2)} = x

    Use subtraction identity to see that ln(18)-ln(2)= ln(\frac{18}{2})=ln(9)
    \frac{ln(54)}{ln(9)} = x

    Now you've solved for x, x=\frac{ln(54)}{ln(9)}\approx 1.82
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  3. #3
    Super Member angel.white's Avatar
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    The second equation will be x \approx -2.63 can you figure out what the exact form is?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    an alternate approach. (i originally thought it could be done without logs, so i tried simplifying to equate like powers, but i had to end up using them at the end)

    3 \cdot 2^x = 18^{x - 1}

    \Rightarrow 3 \cdot 2^x = \left( 3^2 \cdot 2 \right)^{x - 1}

    \Rightarrow 3 \cdot 2^x = 3^{2x - 2} \cdot 2^{x - 1}

    \Rightarrow 3 \cdot 2^x = \frac 1{18} 3^{2x} \cdot 2^x .........2^x is never zero so we can divide by it

    \Rightarrow 3 = \frac {3^{2x}}{18}

    \Rightarrow 3^{2x} = 3 \cdot 18 = 54

    here is where i said, "Oh, darn it! I need logs!"

    \Rightarrow 2x \ln 3 = \ln 54

    \Rightarrow x \ln 9 = \ln 54

    \Rightarrow x = \frac {\ln 54}{\ln 9} \approx 1.82 as angel.white said
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Jhevon View Post
    an alternate approach. (i originally thought it could be done without logs, so i tried simplifying to equate like powers, but i had to end up using them at the end)

    3 \cdot 2^x = 18^{x - 1}

    \Rightarrow 3 \cdot 2^x = \left( 3^2 \cdot 2 \right)^{x - 1}

    \Rightarrow 3 \cdot 2^x = 3^{2x - 2} \cdot 2^{x - 1}

    \Rightarrow 3 \cdot 2^x = \frac 1{18} 3^{2x} \cdot 2^x .........2^x is never zero so we can divide by it

    \Rightarrow 3 = \frac {3^{2x}}{18}

    \Rightarrow 3^{2x} = 3 \cdot 18 = 54

    here is where i said, "Oh, darn it! I need logs!"

    \Rightarrow 2x \ln 3 = \ln 54

    \Rightarrow x \ln 9 = \ln 54

    \Rightarrow x = \frac {\ln 54}{\ln 9} \approx 1.82 as angel.white said
    I'm impressed that you thought of that, even if I seen that, I think I would have gotten too lost after just a few steps, and given up lol, gj.
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  6. #6
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    Quote Originally Posted by imthatgirl View Post
    2) 2(27)^x = 9^(x+1)
    For this one, use the same strategy Angel White and Jhevon did, by taking the log of both sides.

    \ln \left[ 2(27)^x \right] = \ln 9^{x+1},

    Then manipulate the equation using what you know about logs.

    \ln 2 + x \ln 27 = (x+1) \ln 9.

    Now no more x's are exponents, so you can just rearrange this equation using more elementary techniques to isolate the x.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    I'm impressed that you thought of that, even if I seen that, I think I would have gotten too lost after just a few steps, and given up lol, gj.
    yeah, crazy people see things sometimes. i'd also do some manipulations for the second question before applying logs as well

    2 \cdot 27^x = 9^{x + 1}

    \Rightarrow 2 \cdot 3^{3x} = 3^{2x + 2} .......3^(3x) is never zero, so we can divide by it

    \Rightarrow 2 = \frac {3^{2x + 2}}{3^{3x}}

    \Rightarrow 2 = 3^{2 - x}

    now apply logs, the algebra of solving for x will be a lot simpler now
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  8. #8
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    Thanks so much!!! and yeah i get it now, but i did the second one and i got a different answer, i actually got

    log(2/9) / log(1/3) = 1.369
    and when i substitute it back in, it worked... but i'll compare answers with someone tomorrow...
    but thanks so much for helping
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by imthatgirl View Post
    Thanks so much!!! and yeah i get it now, but i did the second one and i got a different answer, i actually got

    log(2/9) / log(1/3) = 1.369
    and when i substitute it back in, it worked... but i'll compare answers with someone tomorrow...
    but thanks so much for helping
    that is the correct answer. what method did you use to arrive at it?
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  10. #10
    Super Member angel.white's Avatar
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    Quote Originally Posted by imthatgirl View Post
    Thanks so much!!! and yeah i get it now, but i did the second one and i got a different answer, i actually got

    log(2/9) / log(1/3) = 1.369
    and when i substitute it back in, it worked... but i'll compare answers with someone tomorrow...
    but thanks so much for helping
    That is correct, my apologies, I transcribed the right side to my paper as 9^{x-1} so I solved for a different equation
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  11. #11
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    I used angel.white's method
    but thanks for the help,

    and no worries angel.white, u explained really well for the first one =D
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