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2. Solve each of the following then check:
a. 2^4x = 4^y+3
b. 9^4x = 27^x-1

User Saju
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Answer:

a. 2^(4x) = 4^(y+3)

We can rewrite 4 as 2^2:

2^(4x) = (2^2)^(y+3)

Using the exponent rule for a power of a power, we get:

2^(4x) = 2^(2(y+3))

Since the bases are the same, the exponents must be equal:

4x = 2(y+3)

4x = 2y + 6

2x = y + 3

y = 2x - 3

Substituting this value of y back into the original equation, we get:

2^(4x) = 4^(y+3)

2^(4x) = 4^(2x)

Taking the square root of both sides:

2^(2x) = 2^(4x/2)

2^(2x) = 2^(2x)

This equation is true for all values of x and y. Therefore, the solution is:

y = 2x - 3

b. 9^(4x) = 27^(x-1)

We can rewrite 27 as 3^3:

9^(4x) = (3^3)^(x-1)

Using the exponent rule for a power of a power, we get:

9^(4x) = 3^(3(x-1))

Simplifying:

3^(2(4x)) = 3^(3(x-1))

Taking the logarithm base 3 of both sides:

2(4x) = 3(x-1)

8x = 3x - 3

5x = -3

x = -3/5

To check, we substitute x = -3/5 back into the original equation:

9^(4(-3/5)) = 27^((-3/5)-1)

9^(-12/5) = 27^(-8/5)

Taking the cube root of both sides:

3^(-4/5) = 3^(-8/5)

This equation is true, so the solution is:

x = -3/5
User Pdc
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