Question

If 4^x=7 then 4^-2x =?

Answer 1

square both sides by \[2\] then invert

Answer 2

since \(\large 4^x=7 \rightarrow x=log_47 \)
then \(\large 4^{-2x}=4^{-2 \cdot (log_47)} \)

simplify....

Answer 3

\[(4^x)^2=7^2\]
\[\frac{ 1 }{ (4^x)^2 }=4^{-2x}\]
so
\[\frac{ 1 }{ 7^2 }\]

Answer 4

if i square both sides ill get 16x^2=49

Answer 5

then apply inversion
rule
\[a^{-1}=\frac{ 1 }{ a }\]

Answer 6

\[\frac{ 1 }{ 16x^{-2} }=\frac{ 1 }{ 49 }\]