Solve for x:
3^(x+1)=2^x
This is what I have so far...
(x+1)log3 = xlog2
...and now I'm stuck. Help plz? Thanks!
that is it, so x+1=x (log2/log3)
x(1-log2/log3)=-1
solve for x
expand the left side ...
xlog3 + log3 = xlog2
xlog3 - xlog2 = -log3
x(log3-log2) = -log3
x = -log3/(log3 - log2)
I got x = appr. - 2.7095
check:
3^(-1.7095) = .15288
2^-2.7095 = .15288 , Ok!
To solve the equation 3^(x+1) = 2^x, you have already made a good start by taking the logarithm of both sides. However, you made a small error in your equation. Let me explain the correct steps to solve the equation:
1. Take the logarithm of both sides using any base. The choice of base depends on personal preference or convenient calculations, but common choices are natural logarithm (base e) or logarithm of base 10.
Taking logarithm base 10, the equation becomes:
log(3^(x+1)) = log(2^x)
2. Use the logarithmic property that states log(a^b) = b * log(a). Apply this property on both sides of the equation:
(x+1) * log(3) = x * log(2)
3. Distribute the logarithm on the left-hand side:
(x * log(3)) + log(3) = x * log(2)
4. Move all terms containing x to one side and all constants to the other side:
x * log(3) - x * log(2) = -log(3)
5. Factor out the common x:
x * (log(3) - log(2)) = -log(3)
6. Simplify the logarithmic terms:
x * log(3/2) = -log(3)
7. Divide both sides of the equation by log(3/2):
x = -log(3) / log(3/2)
Now, you can use a calculator to find the approximate value of x.