Question

6^2x-1 /3^x-1 = 2^x+1
Logarithm problem. I'm solving by taking the log of both sides, then combing and simplifying the all logs as far as possible. I got 2 different answers when I solved it myself and am unsure :/

Answer 1

One sec, let me try.

Answer 2

\[6^{2x-1}~or~6^{2x}-1\]

Answer 3

the 1st one

Answer 4

You can test your answer in the original equation to see if the equation is true.

Answer 5

\[ \frac{6^{2x-1} }{3^{x-1 }}= 2^{x+1} \]

you can take the ln of both sides (or log_10 if you like)
and solve for x.

or you can write 6 as 2*3, and first simplify things:
\[ \frac{(2\cdot 3)^{2x-1} }{3^{x-1 }}= 2^{x+1} \\
2^{2x-1}\frac{3^{2x-1} }{3^{x-1 }}= 2^{x+1}
\]
when you have the same base, you can add or subtract exponents.
3^(2x-1) / 3^(x-1) becomes 3^x (because 2x-1 - (x-1) = x)
\[ 2^{2x-1} 3^x = 2^{x+1} \\
\frac{2^{2x-1}}{2^{x+1} } 3^x = 1\\
2^{x-2} 3^x = 1
\]
write this as
\[ 2^x 2^{-2} 3^x = 1 \\
\left( 2\cdot 3\right)^x \frac{1}{4} = 1 \\
6^x =4
\]
now take the ln of both sides
\[ x \ln(6) = \ln(4) \\ x= \frac{\ln(4)}{\ln(6) }\]

Answer 6

o so sorry and thank you so much! I thought everyone gave up on this one. Thank you for the tips too! :)

Answer 7

the other way is
\[ \ln\left(\frac{6^{2x-1} }{3^{x-1 }}\right)= \ln\left(2^{x+1}\right)\\
\ln(6^{2x-1}) - \ln( 3^{x-1 }) = \ln\left(2^{x+1}\right)\\
(2x-1) \ln(6) - (x-1)\ln(3) = (x+1)\ln(2)
\]
now distribute and collect terms:
\[ 2 \ln(6) x - \ln(6) - \ln(3) x + \ln(3) - \ln(2) x = \ln(2) \\
(2 \ln(6) - \ln(3) - \ln(2)) x = \ln(2)- \ln(3)+ \ln(6)\\
x= \frac{\ln(2)- \ln(3)+ \ln(6)}{2 \ln(6) - \ln(2) - \ln(3)}
\]

Answer 8

Yes my teacher taught us this last way and it's so confusing! I'm going to try to learn it your way :)