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Solve the exponential equation for x. {:[((1)/(4))^(9x-5)=32^(x+8)],[x=◻]:}

Solve the exponential equation for x x .\newline(14)9x−5=32x+8x=□ \begin{array}{l} \left(\frac{1}{4}\right)^{9 x-5}=32^{x+8} \\ x=\square \end{array}

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Full solution

Q. Solve the exponential equation for x x .\newline(14)9x−5=32x+8x=□ \begin{array}{l} \left(\frac{1}{4}\right)^{9 x-5}=32^{x+8} \\ x=\square \end{array}
  1. Rewrite using base 22: Rewrite the left side of the equation using base 22: \newline(14)(9x−5)(\frac{1}{4})^{(9x-5)} = (2−2)(9x−5)(2^{-2})^{(9x-5)} = 2−2∗(9x−5)2^{-2*(9x-5)} = 2−18x+102^{-18x+10}.
  2. Equate exponents: Rewrite the right side of the equation using base 22: (32(x+8))=(25)(x+8)=25∗(x+8)=25x+40.(32^{(x+8)}) = (2^5)^{(x+8)} = 2^{5*(x+8)} = 2^{5x+40}.
  3. Solve for x: Now that both sides of the equation are expressed in terms of the same base, we can equate the exponents: \newline−18x+10=5x+40-18x + 10 = 5x + 40.
  4. Divide to find xx: Solve for xx by moving all terms involving xx to one side and constants to the other side:\newline−18x−5x=40−10-18x - 5x = 40 - 10,\newline−23x=30-23x = 30.
  5. Divide to find x: Solve for x by moving all terms involving x to one side and constants to the other side:\newline−18x−5x=40−10-18x - 5x = 40 - 10,\newline−23x=30-23x = 30.Divide both sides by −23-23 to solve for x:\newlinex=30/(−23)x = 30 / (-23),\newlinex=−30/23x = -30/23.

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