Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 weeks ago

How would I solve this exponential equation?

2^x = 7^(x^2)

the (x^2) in the 7 is throwing me off

thanks!

7 Answers

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  • 2 weeks ago

    2^x = 7^(x^2)

    x log 2 = x² log 7

    x = log 2 / log 7

    x = 0.30103 / 0.8451

    x =0.3562

  • 2 weeks ago

    x = (ln(2) +/- sqrt(ln^2(2) + 8*i*pi*n*ln(7))) / (2*ln(7)), for any integer n

  • ?
    Lv 7
    2 weeks ago

    2^x = 7^(x^2)

    log₂ 2^x = log₂ 7^(x^2)

    x log₂ 2 = (x^2) log₂ 7

    x (1) = (x^2) log₂ 7

    x

    ---- = log₂ 7

    x^2

    1

    -- = log₂ 7

    x

    x log₂ 7 = 1

    ...........1

    x = -----------

    .......log₂ 7

    .....// Change log base

    ...................log 7

    .....log₂ 7 = -------

    ...................log 2

    x = 1 / (log 7 / log 2)

    x = log 2 / log 7

    x = 0.356207...............ANS

  • 2 weeks ago

    x*log(2) = (x^2)*log(7).

    One solution is x = 0.

    The other solution is the solution to

    1*log(2) = x*log(7) =>

    x = log(2)/log(7) = around 0.3562.

  • How do you think about the answers? You can sign in to vote the answer.
  • 2^x = 7^(x^2)

    x * ln(2) = (x^2) * ln(7)

    0 = ln(7) * x^2 - ln(2) * x

    0 = x * (ln(7) * x - ln(2))

    x = 0

    ln(7) * x - ln(2) = 0

    ln(7) * x = ln(2)

    x = ln(2) / ln(7)

    x = 0 , ln(2)/ln(7)

    There you go.

  • hfshaw
    Lv 7
    2 weeks ago

    You have:

    2^x = 7^(x²)

    Take the logarithm of both sides:

    x*ln(2)= (x²)*ln(7)

    By inspection, one solution is x = 0. Having accounted for this solution, we can divide both sides by x:

    ln(2) = x*ln(7)

    x = ln(2)/ln(7)

    So the solutions are x = 0, ln(2)/ln(7).

  • 2 weeks ago

    unable to solve based on your input     

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