# How do you solve the equation #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.

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

Let's apply that to our equation.

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

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

And factor the x's

We can use our log rules to simplify.

Then solve for x.

If we further simplify, we find that

Due to the obscure rule

So finally, you get

By signing up, you agree to our Terms of Service and Privacy Policy

To solve the equation (3^{x-1} \leq 2^{x-7}), you can use the following steps:

- Take the natural logarithm (ln) of both sides of the inequality to remove the exponents.
- Use properties of logarithms to simplify the expression.
- Solve the resulting inequality for (x).
- Check for any restrictions on (x) that may result from the original equation.

Applying these steps:

- (\ln(3^{x-1}) \leq \ln(2^{x-7}))
- ( (x-1) \ln(3) \leq (x-7) \ln(2) )
- Distribute the logarithms: ( x\ln(3) - \ln(3) \leq x\ln(2) - 7\ln(2) )
- Group (x) terms: ( x(\ln(3) - \ln(2)) \leq -7\ln(2) + \ln(3) )
- Solve for (x): ( x \geq \frac{-7\ln(2) + \ln(3)}{\ln(3) - \ln(2)} )

Thus, the solution for (x) is ( x \geq \frac{-7\ln(2) + \ln(3)}{\ln(3) - \ln(2)} ). Make sure to check for any additional restrictions on (x) that may arise from the original equation, such as when taking the logarithm of a negative number. However, in this case, since both (3^{x-1}) and (2^{x-7}) are positive for real (x), no additional restrictions are necessary.

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7