Home > Precalculus > Logarithmic Models

How do you solve # 9^(2x)=27^(x-1)#?

Answer 1

Elijah Allen

#color(blue)(x=-3#

#9^(2x)=27^(x−1#

We know that #9=3^2# #27=3^3#

So,

#3^(2(2x))=3^(3(x−1)#

#3^(4x)=3^(3x−3)#

Now as bases are equal we can equate exponents

#4x=3x-3# #color(blue)(x=-3#

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

Answer 2

Greyson Brown

To solve the equation 9^(2x) = 27^(x-1), we can rewrite 9 as 3^2 and 27 as 3^3, because both are powers of 3. This gives us the equation (3^2)^(2x) = (3^3)^(x-1). Simplifying both sides, we get 3^(4x) = 3^(3x - 3). Since the bases are the same, we can equate the exponents, so 4x = 3x - 3. Solving for x, we find x = -3.

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

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.