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Solve. 7^(x – 5) = 7^(4x + 7)
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\[ \huge 7x-5=4x+7 \] \[\huge 7x-4x = 7+5 =12\] \[\huge 3x =12\] \[\huge x=\frac { 12 }{ 3 } =4 \]
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lol my pleasure
The question should be \[\huge 7^{x-5}=7^{4 x+7}\] \[\huge \begin{array}{l} \mathbb{Same~ base} \\ x-5=4 x+7 \\ \text{Add }(5-4 x)\text{ both sides} \\ -3 x=12 \\ \text{Divide both sides by }-3 \\ x=-4 \\\end{array}\]
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