Question

for each of the following equations, solve for the variable. answers should be integers or fractions. $7^{4y} = 49^{(y - 8)}$ $y = \square$ $64^{(7y - 8)} = 32^{(9y - 11)}$ $y = \square$

Explanation:

First Equation: \( 7^{4y} = 49^{(y - 8)} \)

Step 1: Express 49 as a power of 7

Since \( 49 = 7^2 \), we can rewrite the right - hand side of the equation. So the equation \( 7^{4y}=49^{(y - 8)} \) becomes \( 7^{4y}=(7^2)^{(y - 8)} \).

Step 2: Use the power - of - a - power rule \((a^m)^n=a^{mn}\)

Applying the power - of - a - power rule to \((7^2)^{(y - 8)}\), we get \( 7^{2(y - 8)}=7^{2y-16} \). Now our equation is \( 7^{4y}=7^{2y - 16} \).

Step 3: Set the exponents equal (since the bases are the same and the base is not 0 or 1)

If \( a^m=a^n \) and \( a>0,a
eq1 \), then \( m = n \). So we set \( 4y=2y - 16 \).

Step 4: Solve for y

Subtract \( 2y \) from both sides of the equation \( 4y=2y - 16 \): \( 4y-2y=2y - 16-2y \), which simplifies to \( 2y=-16 \). Then divide both sides by 2: \( y=\frac{-16}{2}=-8 \).

Second Equation: \( 64^{(7y - 8)}=32^{(9y - 11)} \)

Step 1: Express 64 and 32 as powers of 2

We know that \( 64 = 2^6 \) and \( 32 = 2^5 \). So we can rewrite the equation as \( (2^6)^{(7y - 8)}=(2^5)^{(9y - 11)} \).

Step 2: Use the power - of - a - power rule \((a^m)^n=a^{mn}\)

Applying the power - of - a - power rule, we have \( 2^{6(7y - 8)}=2^{5(9y - 11)} \). Simplify the exponents: \( 2^{42y-48}=2^{45y - 55} \).

Step 3: Set the exponents equal (since the bases are the same and \( a = 2>0,a

eq1 \))
Set \( 42y-48 = 45y-55 \).

Step 4: Solve for y

Subtract \( 42y \) from both sides: \( 42y-48-42y=45y - 55-42y \), which gives \( - 48 = 3y-55 \). Then add 55 to both sides: \( -48 + 55=3y-55 + 55 \), so \( 7 = 3y \). Divide both sides by 3: \( y=\frac{7}{3} \).

Answer:

For the equation \( 7^{4y}=49^{(y - 8)} \), \( y=-8 \); for the equation \( 64^{(7y - 8)}=32^{(9y - 11)} \), \( y = \frac{7}{3} \)