Question

Solve for "x" 2(3^x)=12^x-1.

Answer 1

Is this it?
\[2\times3^x=12^{x-1}\]?

Answer 2

yes

Answer 3

this one takes a minute
gimmick is to write
\[12=3\times 4\] so
\[12^{x-1}=3^{x-1}4^{x-1}\] then divide both sides of
\[2\times 3^x=3^{x-1}4^{x-1}\] by \(3^{x-1}\)

Answer 4

maybe @Ahsome has a slicker way, not sure if that method is best

Answer 5

maybe adding a log will help?

Answer 6

Hmmm..

Answer 7

Would we use logarithim?

Answer 8

i believe so

Answer 9

once you ave
\[6=4^{x-1}\] you would use the change of base formula so yes, logarithms for sure

Answer 10

but that is the last step, not the first step

Answer 11

how did you get there?

Answer 12

According to wolframalpha, \[x\approx2.2925\]

Answer 13

i wrote the steps above
take a look and see if you have any questions about how i got to
\[6=4^{x-1}\]

Answer 14

\[2×3^x=12^{x−1}\]
\[3^x=\frac{12^{x−1}}{2}\]
\[log_3{\frac{12^{x−1}}{2}}=x\]

Answer 15

Hmmm

Answer 16

which you and also write as
\[24=4^x\] making
\[x=\frac{\ln(24)}{\ln(4)}\]

Answer 17

yeah i get that

Answer 18

Wait a minute is:
\[2×3^x=12^{x−1}\]
The same as:
\[2×3^x=12^{x}\div12^1?\]

Answer 19

How did you go from 12=3×4 to 12^x-1=3^x-14^x-1

Answer 20

Me, or @satellite73?

Answer 21

\[2×3^x=12^{x−1}\]
\[2×3^x=12^x÷12^1\]
\[12(2\times3^x)=12^x\]
\[log_{12}{12(2\times3^x)}=x\]