2^(3x + 1) = 3^(x − 2)
(a) Find the exact solution of the exponential equation in terms of logarithms.
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 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?
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
To find the exact solution of the exponential equation 2^(3x + 1) = 3^(x − 2) in terms of logarithms, we can take the logarithm of both sides of the equation.
First, let's take the logarithm of both sides using any base, such as the natural logarithm (ln) or common logarithm (log):
ln(2^(3x + 1)) = ln(3^(x − 2))
Using the logarithmic property for exponents, we can bring down the exponent as a coefficient:
(3x + 1) ln(2) = (x - 2) ln(3)
Now, let's isolate the variable x. We can distribute the logarithms:
3x ln(2) + ln(2) = x ln(3) - 2 ln(3)
Next, let's gather the x terms on one side of the equation and the constant terms on the other side:
3x ln(2) - x ln(3) = - ln(2) - 2 ln(3)
Now, factor out the x on the left side:
x (3 ln(2) - ln(3)) = - ln(2) - 2 ln(3)
Finally, divide both sides by (3 ln(2) - ln(3)) to solve for x:
x = (- ln(2) - 2 ln(3)) / (3 ln(2) - ln(3))
This is the exact solution of the exponential equation in terms of logarithms.