Question

3^2x=81^x+9

Answer 1

Do i use a different rule other than power rule to solve the other side?

Answer 2

\[3^{2x} = 81^x+9\]
can you first make all the bases 3?

Answer 3

wait there would be no solution

do you mean 81^{x+9)?

Answer 4

\[3^{2x}=81^{x+9}\\3^{2x}=3^{(4)({x+9})}\]

Answer 5

you probably know how to solve it from there

Answer 6

Sorry, i didnt see you reply, but how did you get that? what rule did you use?

Answer 7

\[3^{2x}=81^{x+9}\\3^2x = 3^4 to ~the~power~of~x+9\\3^{2x}=3^{(4)({x+9})}\]

Answer 8

a^b^c=a^(bc)

Answer 9

ok so first 3^2x=81^x+9
=3^2x=3^4<

Answer 10

3^2x does not equal 3^4

Answer 11

would that be a base rule or exponent rule?

Answer 12

i meant 3^4=81

Answer 13

it would be the power rule...
base rule is for logarithms.. exponent rule is for something else

Answer 14

sorry, thats confusing

Answer 15

81^x+9 is 3^4 ^ x+9 so by the power rule it equals 3^(4(x+9))

Answer 16

but you changed bases,?

Answer 17

yeah latex doesn't allow me to do double exponents i think

Answer 18

i don't think the names of rules really matter

Answer 19

it's what to actually do

Answer 20

yes

Answer 21

but i alwyas forget that rule

Answer 22

thank you thats the rule i was looking for

Answer 23

yeah I guess it's called power rule for exponents