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Solve the exponential equation for x. {:[3^(7x+4)=((1)/(27))^(x-3)],[x=◻]:}

Solve the exponential equation for x x .\newline37x+4=(127)x−3x=□ \begin{array}{l} 3^{7 x+4}=\left(\frac{1}{27}\right)^{x-3} \\ x=\square \end{array}

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Q. Solve the exponential equation for x x .\newline37x+4=(127)x−3x=□ \begin{array}{l} 3^{7 x+4}=\left(\frac{1}{27}\right)^{x-3} \\ x=\square \end{array}
  1. Given exponential equation: We are given the exponential equation:\newline37x+4=(127)x−33^{7x+4} = \left(\frac{1}{27}\right)^{x-3}\newlineFirst, we recognize that 2727 is a power of 33, specifically 27=3327 = 3^3. We can rewrite the right side of the equation using this fact.
  2. Rewriting the right side: Rewrite the right side of the equation using 27=3327 = 3^3:\newline(127)x−3=(133)x−3=3−3(x−3)\left(\frac{1}{27}\right)^{x-3} = \left(\frac{1}{3^3}\right)^{x-3} = 3^{-3(x-3)}\newlineNow the equation is:\newline37x+4=3−3(x−3)3^{7x+4} = 3^{-3(x-3)}
  3. Setting the exponents equal: Since the bases are the same, we can set the exponents equal to each other:\newline7x+4=−3(x−3)7x + 4 = -3(x - 3)\newlineNow we will solve for xx.
  4. Solving for x: Distribute the −3-3 on the right side of the equation:\newline7x+4=−3x+97x + 4 = -3x + 9
  5. Distributing −3-3: Add 3x3x to both sides to get all the xx-terms on one side:\newline7x+3x+4=97x + 3x + 4 = 9\newline10x+4=910x + 4 = 9
  6. Combining like terms: Subtract 44 from both sides to isolate the xx-term:\newline10x=9−410x = 9 - 4\newline10x=510x = 5
  7. Isolating the x-term: Divide both sides by 1010 to solve for xx:\newlinex=510x = \frac{5}{10}\newlinex=12x = \frac{1}{2}

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