Math  /  Algebra

QuestionSolve the exporential equation. Expresi irrational solutions as decimals correct to the vearest thovianath. 5x2=32x5^{x-2}=3^{2 x}
Select the correct choice A. The solution set is }\} B. empty set

Studdy Solution

STEP 1

What is this asking? We need to find the value of xx that makes 5x25^{x-2} equal to 32x3^{2x}, and if the answer isn't a nice whole number, we'll round it to the nearest thousandth. Watch out! Exponent rules are key here, so let's make sure we don't mix them up!
Also, remember to round carefully at the end.

STEP 2

1. Apply Natural Logarithm
2. Expand Using Logarithm Rules
3. Isolate *x*
4. Calculate and Round

STEP 3

We **start** with our equation: 5x2=32x5^{x-2} = 3^{2x} Now, to bring those exponents down, we'll take the natural logarithm (ln) of both sides.
Remember, what we do to one side, we *must* do to the other! ln(5x2)=ln(32x)\ln(5^{x-2}) = \ln(3^{2x}) Why is this a good idea?
Because it lets us use our awesome logarithm power rule in the next step!

STEP 4

Remember that handy logarithm power rule?
It says ln(ab)=bln(a)\ln(a^b) = b \cdot \ln(a).
Let's use it! (x2)ln(5)=2xln(3)(x-2) \cdot \ln(5) = 2x \cdot \ln(3) Now, those exponents aren't stuck up there anymore, making things much easier to manage.

STEP 5

Let's distribute ln(5)\ln(5) on the left side: xln(5)2ln(5)=2xln(3)x \cdot \ln(5) - 2 \cdot \ln(5) = 2x \cdot \ln(3) This helps us separate the terms with xx from the constant terms.

STEP 6

We want to get all the terms with xx on one side and everything else on the other.
Let's subtract xln(5)x \cdot \ln(5) from both sides: 2ln(5)=2xln(3)xln(5)-2 \cdot \ln(5) = 2x \cdot \ln(3) - x \cdot \ln(5) This groups our xx terms together, making it easier to solve for xx.

STEP 7

Now, we can factor out xx from the right side: 2ln(5)=x(2ln(3)ln(5))-2 \cdot \ln(5) = x \cdot (2 \cdot \ln(3) - \ln(5)) This isolates xx, which is exactly what we want!

STEP 8

Finally, we can divide both sides by (2ln(3)ln(5))(2 \cdot \ln(3) - \ln(5)) to solve for xx: x=2ln(5)2ln(3)ln(5)x = \frac{-2 \cdot \ln(5)}{2 \cdot \ln(3) - \ln(5)} We've got xx all by itself now!

STEP 9

Using a calculator, we find: ln(5)1.6094\ln(5) \approx 1.6094 ln(3)1.0986\ln(3) \approx 1.0986So, x21.609421.09861.60943.21882.19721.60943.21880.58785.4781x \approx \frac{-2 \cdot 1.6094}{2 \cdot 1.0986 - 1.6094} \approx \frac{-3.2188}{2.1972 - 1.6094} \approx \frac{-3.2188}{0.5878} \approx -5.4781

STEP 10

Rounding to the nearest thousandth gives us: x5.478x \approx -5.478

STEP 11

The solution set is {5.478}\{-5.478\}.
So the correct choice is A.

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