Question

Can someone please answer this question,

4^3x+1 = 7^x

thanks.

Answer 1

Are you asking for solving the solution \[ 4^{3x}+1=7^x \]
? Because your notation is somewhat ambiguous.

Answer 2

*solving the equation

Answer 3

yes please

Answer 4

actually the +1 is part of the exponent

Answer 5

Okay because the other one has no real solutions ;).

Answer 6

The solution is \[ -2\,{\frac {\ln \left( 2 \right) }{6\,\ln \left( 2 \right) -\ln
\left( 7 \right) }} \]

Answer 7

oh lol

Answer 8

Oh ic, any chance you could show me your steps?

Answer 9

Yes

Answer 10

Yea thats how to do it, but you messed up somewhere I guess.

Answer 11

The main idea is take the log on both sides and then use logarithm laws.

Answer 12

Alright thanks to you both for your help, I appreciate it and the answer says -ln4 / 3ln4 - ln7

Answer 13

\[\Large 4^{3x+1} = 7^{x}\]


\[\Large \ln(4^{3x+1}) = \ln(7^{x})\]


\[\Large (3x+1)\ln(4) = x\ln(7)\]


\[\Large 3x\ln(4)+\ln(4) = x\ln(7)\]


\[\Large \ln(4) = x\ln(7)-3x\ln(4)\]


\[\Large \ln(4) = x\left(\ln(7)-3\ln(4)\right)\]


\[\Large \frac{\ln(4)}{\ln(7)-3\ln(4)}=x\]


\[\Large x=\frac{\ln(4)}{\ln(7)-3\ln(4)}\]

Answer 14

something happen to your 3 jim

Answer 15

yep it ran away

Answer 16

ok you got it lol

Answer 17

Yea my answer was the same, but it looks different because I let a CAS do it :D

Answer 18

oh lol, thanks for the help!

Answer 19

to match what the book has, follow the steps given below



\[\Large x=\frac{\ln(4)}{\ln(7)-3\ln(4)}\]


\[\Large x=\frac{\ln(4)}{-3\ln(4)+\ln(7)}\]


\[\Large x=\frac{\ln(4)}{-(3\ln(4)-\ln(7))}\]


\[\Large x=\frac{-\ln(4)}{3\ln(4)-ln(7)}\]

Answer 20

either way, it's the same answer

Answer 21

Oh alright, makes sense , thanks

Answer 22

those threes are tricky