Question

find the value of x so that [7/5]^8 × [49/25]^-2 = [7/5]^2x
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Answer 1

Answer:

$\large \underline{ \sf{Question:-}}$

$\tt \large \rightarrow{ (\frac{7}{5} })^{8} \times { (\frac{49}{25}) }^{ - 2} = { (\frac{7}{5} })^{2x}$

$\sf \large \underline{To \: \: Find:-}$

• We have to find the value of x

$\sf \large \underline{Solution:-}$

$\implies \tt { (\frac{7}{5} })^{8} \times { (\frac{49}{25} })^{ - 2} = {( \frac{7}{5} })^{2x}$

$\implies \tt \ { (\frac{7}{5} })^{8} \times { (\frac{ {7}^{2} }{ {5}^{2} } })^{ - 2} = { \frac{7}{5} }^{2x}$

$\implies \tt { (\frac{7}{5} })^{8} \times { (\frac{7}{5} })^{ - 2 \times 2} = {( \frac{7}{5} })^{2x}$

$\implies \tt { (\frac{7}{5} })^{8} \times { (\frac{7}{5} })^{ - 4} = {( \frac{7}{5} })^{2x}$

As we know that if bases are equal then we equate powers,

$\implies \sf8 - 4 = 2x$

$\implies \sf4 = 2x$

$\sf \implies x = \frac{4}{2} = \red2$

$\sf So, Value \: \: of \: \: \red{ x = 2}$