Question

6^(1-x) = 2^(3x+1)

Answer 1

3 * 2^(1-x) = 2^(3x+1)

Answer 2

would that be how I would approach it?

Answer 3

yeah. when you have the same base you can do the next step

Answer 4

but since we break 6 into 3*2, shouldn't the exponent also be put on the 3 too? I've try that but I get stuck

Answer 5

yeah you're right

Answer 6

so it will be 3^(1-x) * 2^(1-x) = 2^(3x+1)

Answer 7

can you use logs for your solution?

Answer 8

ok, but when i divide over...I eventually get 3=3^x * 2^(4x)
and yes, but i'm not sure how i could use log in this situation

Answer 9

or in other form 3^(1-x) = 2^4x

Answer 10

well you got the answer! 3^x * 2^(4x) = 3

Answer 11

but i'm looking for x

Answer 12

i'm trying to get x by itself

Answer 13

then you need logs to do that

Answer 14

I have 2 different base numbers on one side though, I don't know how I can change that to log

Answer 15

log 3 = x log 3 * 4x log 2

Answer 16

\[(1-x)\ln(5)=(3x+1)\ln(2)\] is a start

Answer 17

actually that 5 should be a 6

Answer 18

ok.

Answer 19

\[\large
6^{1-x} = 2^{3x+1} \\
\text{Take log on both sides:} \\ \large
(1-x)\log(6) = (3x+1)\log(2) \\ \large
\log(6) - x\log(6) = 3\log(2)x + \log(2) \\ \large
\log(6) - \log(2) = 3\log(2)x + x\log(6) = x\{3\log(2) + \log(6)\} \\ \large
\log{\frac 62} = x\{\log(2)^3 + \log(6)\} \\ \large
\log(3) = x\{\log(8) + \log(6)\} = x * \log(48) \\ \large
x = \frac{\log(3)}{\log(48)}
\]

Answer 20

ok. thanks

Answer 21

you are welcome.