3^(4x+12) = 5^(3x+y)

BTGModeratorVI wrote: ↑

Thu Jun 18, 2020 5:41 am

For integers x and y, 3^(4x+12) = 5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

Answer: D
Source: Veritas Prep

The key word in this question is INTEGERS

Notice that, if x is an integer, then 4x+12 is an integer, which means 3^(4x+12) will equal the product of a bunch of 3's
Likewise, if x and y are integers, then 3x+y is an integer, which means 5^(3x+y) will equal the product of a bunch of 5's
Given these conditions, it seems impossible that 3^(4x+12) could ever equal 5^(3x+y)
HOWEVER, if the exponents 4x+12 and 3x + y both equal ZERO, then we get 3^0 and 5^0, and both of these evaluate to equal 1 - PERFECT!

So, let 4x+12 = 0 and let 3x+y = 0
Now we'll solve this system of equations for x and y.

First, if 4x+12 = 0, then x = -3
If x = -3, then we can take 3x+y = 0 and replace x with -3 to get: 3(-3) + y = 0
Simplify: -9 + y = 0
Solve: y = 9

So, x = -3 and y = 9, is a solution to the equation 3^(4x+12) = 5^(3x+y)

Answer: D

Cheers,
Brent

BTGModeratorVI wrote: ↑

Thu Jun 18, 2020 5:41 am

For integers x and y, 3^(4x+12) = 5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

Answer: D

Solution:

In order for a power in base 3 to equal a power in base 5, the exponents of each base must equal zero. In other words, 3^0 = 1 and 5^0 = 1, and so 3^0 = 5^0. Since prime factorization of integers is unique, this is the only way that the original equation can be satisfied. Thus, since each exponent must equal 0, we have:
4x + 12 = 0

4x = -12

x = -3

Solving for y, we have:

3(-3) + y = 0

-9 + y = 0

y = 9

Answer: D