Solve for X. Leave answers in logarithmic forms.
3^(2-x) = 2^x
Thanks.
take the logs of both sides
log (3^(2-x)) = log (2^x)
(2-x)log3 = xlog2
2log3 - xlog3 = xlog2
2log3 = xlog2 + xlog3
2log3 = x(log2 + log3)
x = 2log3/(log2 + log3)
or x = 2log3/log6
Ok, I'm kind of confused.
A log's base typically 10 right? Why would you write log 3^(2-x)? Shouldn't it be log(sub3) 2-x) ?
To solve this equation, we need to use the properties of logarithms. Specifically, we will take the logarithm of both sides of the equation.
First, let's take the logarithm base 3 of both sides:
log₃(3^(2-x)) = log₃(2^x)
Using the logarithm property logₐ(a^b) = b * logₐ(a), we can simplify the left side:
(2 - x) * log₃(3) = x * log₃(2)
Since logₐ(a) = 1 for any base a, we have:
(2 - x) * 1 = x * log₃(2)
Simplifying further:
2 - x = x * log₃(2)
Now, let's isolate the variable x. Move the x term to one side of the equation:
2 = x * log₃(2) + x
Factoring out the variable x on the right side:
2 = x * (log₃(2) + 1)
Divide both sides of the equation by (log₃(2) + 1):
2 / (log₃(2) + 1) = x
Therefore, the solution to the equation is:
x = 2 / (log₃(2) + 1)
In logarithmic form, we can express the answer as:
log₃(2^(2 / (log₃(2) + 1))) = x