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Math Help - Intersection of 2 log functions

  1. #1
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    Intersection of 2 log functions

    So the log functions given are y=3(5^(2x)) and y=6(4^(3x)) .

    We are asked to find their point of intersection. To do this, I set each one to equal the other, 3(5^(2x))=6(4^(3x)) , and solve to find the x-value that the two meet at, then I'll sub that in later to find the y-value....


    But how do I solve this: 3(5^(2x))=6(4^(3x)) ....... I can't seem to figure that one out. We are doing logarithmic equations practice... We know the product, quotient and power laws of logs.


    Thanks!
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  2. #2
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    Quote Originally Posted by mike_302 View Post
    So the log functions given are y=3(5^(2x)) and y=6(4^(3x)) .

    We are asked to find their point of intersection. To do this, I set each one to equal the other, 3(5^(2x))=6(4^(3x)) , and solve to find the x-value that the two meet at, then I'll sub that in later to find the y-value....


    But how do I solve this: 3(5^(2x))=6(4^(3x)) ....... I can't seem to figure that one out. We are doing logarithmic equations practice... We know the product, quotient and power laws of logs.


    Thanks!
    Sorry if I'm misunderstanding this, but what exactly are the logs?

    Is it \log_{3} 5^{2x} ?
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  3. #3
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    3(5^{2x})=6(4^{3x})

    If we divide by 3 and 4^{3x}, we get:

    \frac{5^{2x}}{4^{3x}} = 2

    \implies \left(\frac{5^2}{4^3}\right)^x = 2

    Can you do it now using what you know with laws of logs?
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  4. #4
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    sorry: Bad wording on my part... Chop Suey's got it though.. I'm just trying to work out how you went from your first step to the last.
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  5. #5
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    Yep, I've figured it out... works out to -0.737 (approx) .

    Thanks!
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  6. #6
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    Quote Originally Posted by mike_302 View Post
    sorry: Bad wording on my part... Chop Suey's got it though.. I'm just trying to work out how you went from your first step to the last.
    One of the properties of exponents are:
    (ab)^m = a^m b^m

    In this case, we have \frac{a^{mn}}{b^{on}} = \left(\frac{a^m}{b^o}\right)^n
    EDIT:
    Quote Originally Posted by mike_302 View Post
    Yep, I've figured it out... works out to -0.737 (approx) .

    Thanks!
    Good job.
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