# How do you solve the equation 3^(x-1)<=2^(x-7)?

May 19, 2017

x ≤ log_(3/2)(3/2^7)

#### Explanation:

3^(x−1)≤2^(x−7)

First, since there are x's in the exponents, we know that we'll be using logs to get those down.

Let's take the natural log of both sides.

ln(3^(x−1))≤ln(2^(x−7))

In log rules, remember that any exponent inside the log can become the log's coefficient. Or, in math language:

log(x^a) ≤ alog(x)

Let's apply that to our equation.

(x-1)ln(3) ≤ (x-7)ln(2)

Now, I'm going to factor both sides, because dividing by x wouldn't be very helpful.

ln(3)x-ln(3) ≤ ln(2)x-7ln(2)

Now let's get x on one side of the equation.

ln(3)x - ln(2)x ≤ ln(3)-7ln(2)

And factor the x's

x[ln(3) - ln(2)] ≤ ln(3)-7ln(2)

We can use our log rules to simplify.

x[ln(3/2)] ≤ ln(3/2^7)

Then solve for x.

x ≤ ln(3/2^7)/[ln(3/2)]

If we further simplify, we find that

$\ln \frac{\frac{3}{2} ^ 7}{\ln \left(\frac{3}{2}\right)} = {\log}_{\frac{3}{2}} \left(\frac{3}{2} ^ 7\right)$

Due to the obscure rule

${\log}_{b} \left(x\right) = {\log}_{a} \frac{x}{\log} _ a \left(b\right)$

So finally, you get

x ≤ log_(3/2)(3/2^7)