Question

How do you solve 49^x = sqrt7 ^(2x+6)?

Answer 1

\[49^x = \sqrt{7^{2x+6}}\]?

Answer 2

No, sorry, i meant 49^x = sqrt(7) ^ 2x+6

Answer 3

\[\sqrt{7}^{2x+6}\]
I guess

Answer 4

49^x=7^x+3 ..by cancelling sqrt with power

Answer 5

x=3

Answer 6

zzr0ck3r how did you eliminate the square root on the 7 and put two there?

Answer 7

\[\sqrt{7}=7^{\frac{1}{2}}\]

Answer 8

aha

Answer 9

\[49^x = \sqrt{7}^{2x+6}=(7^{\frac{1}{2}})^{2x+6}\\7^{2x}=7^{x+3}\\take \space \log_7\space of\space both\space sides\\2x=x+3\\x=3\]

Answer 10

I understand it now...
Thank you so much!!

Answer 11

np