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[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle 49^{5x+2} = \left(\frac{1}{7}\right)^{11 - x}\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle 49^{5x+2} = \left(\frac{1}{7}\right)^{11 - x}\)

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Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle \frac{1}{7} = 7^{-1}\)

Then,

\(\displaystyle 49^{5x+2} = 7^{x - 11}\)

 

[logistic_guy]

\(\displaystyle 49^{5x+2} = 7^{x - 11}\)

\(\displaystyle 49 = 7 \times 7 = 7^2\)

Then,

\(\displaystyle 7^{10x+4} = 7^{x - 11}\)

 

[logistic_guy]

\(\displaystyle 7^{10x+4} = 7^{x - 11}\)

Since the base is equal, then we have:

\(\displaystyle 10x+4 = x - 11\)

Solve for \(\displaystyle x\).

\(\displaystyle 10x - x = -11 - 4\)

\(\displaystyle 9x = -15\)

\(\displaystyle x = -\frac{15}{9} = \textcolor{blue}{-\frac{5}{3}}\)