3^x+1=6^4-x
If your answer has logarithms use parenthesis for the numerator and the denominator,
for example: x= (ln3-2ln4)/(-2ln3+ln4)
If the problem is: 3x + 1 = 64 - x
---> ln( 3x + 1 ) = ln( 64 - x )
---> (x + 1)ln(3) = (4 - x)ln(6)
---> x·ln(3) + ln(3) = 4·ln(6) - x·ln(6)
---> x·ln(3) + x·ln(6) = 4·ln(6) - ln(3)
---> x[ ln(3) + ln(6) ] = 4·ln(6) - ln(3)
---> x = [ 4·ln(6) - ln(3) ] / [ ln(3) + ln(6) ]
If the problem is: 3x + 1 = 64 - x
---> ln( 3x + 1 ) = ln( 64 - x )
---> (x + 1)ln(3) = (4 - x)ln(6)
---> x·ln(3) + ln(3) = 4·ln(6) - x·ln(6)
---> x·ln(3) + x·ln(6) = 4·ln(6) - ln(3)
---> x[ ln(3) + ln(6) ] = 4·ln(6) - ln(3)
---> x = [ 4·ln(6) - ln(3) ] / [ ln(3) + ln(6) ]