How do you solve the equation #3^(x-1)

Question

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

Elias Ackerman · Accepted Answer

#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)#

Greyson Brown · Answer

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

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

Applying these steps:

1. $\ln(3^{x-1}) \leq \ln(2^{x-7})$
2. $ (x-1) \ln(3) \leq (x-7) \ln(2) $
3. Distribute the logarithms: $ x\ln(3) - \ln(3) \leq x\ln(2) - 7\ln(2) $
4. Group $x$ terms: $ x(\ln(3) - \ln(2)) \leq -7\ln(2) + \ln(3) $
5. 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.

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

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