take logs of both sides and use log rules
(3x+1)log2 = (x-2)log3
3log2 x + log2 = log3 x - 2log3
x(3log2 - log3) = -2log3 - lo2
x = (2log3 + log2)/(log3 - 3log2)
2^(3x + 1) = 3^(x − 2)
(a) Find the exact solution of the exponential equation in terms of logarithms.
3 answers
2 log(3x+1) = 3 log(x-2)
Get this step by using the law of logs regarding exponents.
log(3x +1) /log (x-2) = 3/2
log (3x + 1 -x +2) = 3/2
log (2x + 3) = 3/2
can you finish from here?
Get this step by using the law of logs regarding exponents.
log(3x +1) /log (x-2) = 3/2
log (3x + 1 -x +2) = 3/2
log (2x + 3) = 3/2
can you finish from here?
John, you misread the question
2^(3x + 1) = 3^(x − 2)
is not the same as
2 log(3x+1) = 3 log(x-2)
then, log(3x +1) /log (x-2) ≠ log (3x + 1 -x +2)
e.g. is log100/log10 = log(100-10) ???
LS = 2/1 = 2
RS = log 90
2^(3x + 1) = 3^(x − 2)
is not the same as
2 log(3x+1) = 3 log(x-2)
then, log(3x +1) /log (x-2) ≠ log (3x + 1 -x +2)
e.g. is log100/log10 = log(100-10) ???
LS = 2/1 = 2
RS = log 90
1 answer