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

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Answer 1 · 2017-05-19T02:43:32

#x ≤ log_(3/2)(3/2^7)#

#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(3/2^7)/[ln(3/2)] = log_(3/2)(3/2^7)#

Due to the obscure rule

#log_b(x) = log_a(x)/log_a(b)#

So finally, you get

#x ≤ log_(3/2)(3/2^7)#