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anonymous

  • 5 years ago

What is 2^5x+3=1/8 in exponential form for logarithmic equations?

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  1. anonymous
    • 5 years ago
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    I'm not sure what you mean, but I think you want the equation in logarithmic form? Because it's already in exponential form.

  2. anonymous
    • 5 years ago
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    Sorry.. I mean to say solve this equation using algebraic procedure. Its a logarithmic equation. So here is the equation once again: 2^5x+3=1/8

  3. anonymous
    • 5 years ago
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    If that's the case, then do the following: 1) subtract 3 to the other side 2^(5x)=(1/8)-3 2) Take the log base 2 of both sides following the rule y = log2(x), 2^y=x \[\log_{2} (2^{5x})=\log_{2}(1/8 - 3)\] 3) The log base 2 cancels the 2 in 2^5x leaving you with 5x \[5x=\log_{2}(1/8-3)\] 4) divide both sides by 5 \[x=\log_2(1/8-3)/5\] This is a logarithmic equation because it has logarithms in it. This is different from an exponential equation which you provided. In fact, it is the inverse.

  4. anonymous
    • 5 years ago
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    But the answer in my textbook is -6/5.. LOL!

  5. anonymous
    • 5 years ago
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    I think I wrote the problem wrong.. its 2^(5x+3)=1/8

  6. anonymous
    • 5 years ago
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    Okay, so the steps are the same. 1) Take the log base 2 of both sides because your base in the exponential function 2^(5x+3) is 2. \[\log_{2}(2^{5x+3})=\log_2(1/8) \] 2) Using the logarithm property \[\log_a(a^{x}) = x\] where x is some function, employ the property on the left hand side of the equation to get\[5x+3=\log_{2}(1/8) \] 3) Another logarithm property states that\[\log_{a}(x/y)=\log_a(x)-\log_a(y) \]where x and y are some function, employ this on the right hand side to get\[5x+3=\log_2(1)-\log_2(8)\] 4) loga(x) = N is equivalent to a^N=x so 2^N=1 when N=0 and 2^N=8 when N=3 giving you\[5x+3=0-3=-3\] 5) Subtract 3 on both sides of the equation and divide both sides of the equation by 5 to get\[x=(-3-3)/5 = -6/5\]

  7. anonymous
    • 5 years ago
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    Thank you so much for your help. :)

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