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Thread: Logarithmic/Exponential Equation

  1. #1
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    Logarithmic/Exponential Equation

    What the heck did I do wrong in the following problem?

    Solve: $\displaystyle 2^{5x-6} = 7$

    Work:
    $\displaystyle log2^{5x-6} = log7$

    $\displaystyle (5x - 6) (log2) = log7$

    $\displaystyle 5x - 6 = (\frac {log7}{log2})$

    $\displaystyle 5x = (\frac {log7}{log2}) + 6$

    $\displaystyle x = \frac {[(\frac {log7}{log2}) + 6]}{5}$


    And when I put both sides to base of 2, I get:

    $\displaystyle log_22^{5x-6} = log_22^{7}$

    $\displaystyle (5x-6)(log_22) = 7(log_22)$

    $\displaystyle (5x-6) = \frac {7(log_22)}{(log_22)}$

    $\displaystyle 5x = 7 + 6$

    $\displaystyle x = \frac{7+6}{5}$

    Textbook Answer: $\displaystyle \frac {log_27 + 6}{5}$
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  2. #2
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    Hello, Macleef!

    Solve: .$\displaystyle 2^{5x-6} = 7$

    Work:

    $\displaystyle \log2^{5x-6} \:= \:\log7$

    $\displaystyle (5x - 6)\log2) \:= \:\log7$

    $\displaystyle 5x - 6 \:= \:\frac{\log7}{\log2}$

    $\displaystyle 5x \:= \:\frac{\log7}{\log2} + 6$

    $\displaystyle x \:= \:\frac{\dfrac{\log7}{\log2} + 6}{5}$ . . . . Right!

    Textbook Answer: $\displaystyle \frac{\log_2\!7 + 6}{5}$ . . . . also correct!
    Using the Base-Change Formula, note that: .$\displaystyle \log_2\!7 \:=\:\frac{\log 7}{\log 2} $


    Your second method was faulty . . .

    Take logs, base 2: .$\displaystyle \log_2\left(2^{5x-6}\right) \:=\:\log_2(7)$

    $\displaystyle \text{Then we have: }\;(5x-6)\underbrace{\log_2(2)}_{\text{This is 1}} \:=\:\log_2(7)$

    Therefore: .$\displaystyle 5x - 6 \:=\:\log_2(7) \quad\Rightarrow\quad x \:=\:\frac{\log_2(7) + 6}{5} $

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