Please solve for X: 3^(2x+1) = 4^x?

Question

Can you please provide step by step detail in addition to solution? Trying to understand concept.

Answer 1

3^(2x+1)=4^x

Take the logarithm of both sides:

(2x+1)lg3=xlg4

Divide both sides, first by a factor x, then by a factor lg3:

(2x+1)/x=lg4/lg3

lg4/lg3 has a known value...

(2x+1)/x is the same as 2+(1/x)

so 2+(1/x) = (lg4/lg3)

and (1/x) = (lg4/lg3) - 2

invert it all:

x/1 = 1/((lg4/lg3) - 2)

Then stick in your values for logs of 3 and 4...

Alternatively, express (lg4/lg3) as the base 3 log of 4 (log(subscript 3)4)...

Answer 2

3^(2x + 1) = 4 ^x ..............1)
Take the log to any base of both sides of eqn 1)
(2x + 1) log 3 = x log 4 ............2)
Expand eqn 2) and bring terms with x to one side
2x log3 - x log4 = - log3 ............3)
Factor out x and divide both sides by factor of x
x (2 log3 - log4) = - log3
Now use your calculator to read the log to base 10 of 3 and 4 and substitute to get value below:
x = -log3/ (2log3 - log4) = -0.7382

Answer 3

log(3)4^x=2x+1
x*log(3)4=2x+1
log(3)4=(2x+1)/x=2+1/x
1/x=log(3)4-2
x=1/[log(3)4-2]

Answer 4

(2x+1)ln3=xln4
2xln3-xln4=-ln3

x=-ln3/(2ln3-ln4)

x=-ln3/ln(9/4)