Question

can some one give me a hand with this?
7^-x * (1/7)^(9x-2)= 49^(x^2-5)

Answer 1

\[\frac{ 1 }{ x }=x^{-1}\]

Answer 2

u know that ???

Answer 3

i know that but i never thought i could solve it like that...............

Answer 4

thank you!

Answer 5

\[(7^{-x})(\frac{ 1 }{ 7 }) ^{9x-2}=49^{x^2-5}\]
is that the given?

Answer 6

yeah, i'm still stuck though

Answer 7

how do u get id of the \[1^{9x-2}\]

Answer 8

rid*

Answer 9

first thing to do, express each term to the base of 7

Answer 10

follow what gorv has pointed out above. express (1/7)^(9x-2) to a base of 7

Answer 11

for the right side, 49 is a perfect square. express it also to the base of 7

Answer 12

\[\frac{ 1 }{ 7 } \rightarrow 7^{-1}\]

Answer 13

\[7^{-10x+2} = 7^{2x^2-10}\]
is that right?

Answer 14

yes it's correct