[T3] The product of two powers is 8¹º and the quotient of the same two powers is 8. What could the two powers be? Verify your work by showing the calculation for both the product and the quotient.

1 answer

Let the two powers be \( a^m \) and \( a^n \).

According to the problem, we have:

  1. The product of the two powers is given as: \[ a^m \cdot a^n = 8^{10} \] Using the property of exponents, this simplifies to: \[ a^{m+n} = 8^{10} \]

  2. The quotient of the same two powers is given as: \[ \frac{a^m}{a^n} = 8 \] This can also be simplified using the property of exponents: \[ a^{m-n} = 8 \]

We can substitute \( 8 \) with \( 2^3 \), giving us: \[ 8^{10} = (2^3)^{10} = 2^{30} \] and \[ 8 = 2^3 \]

Now, we rewrite our equations in terms of base 2:

  1. From \( a^{m+n} = 2^{30} \), we can take logarithms or assume \( a = 2 \), leading to: \[ 2^{m+n} = 2^{30} \implies m+n = 30 \]

  2. From \( a^{m-n} = 2^3 \): \[ 2^{m-n} = 2^3 \implies m-n = 3 \]

We now have a system of equations: \[ \begin{align*} m + n &= 30 \quad \text{(1)} \ m - n &= 3 \quad \text{(2)} \end{align*} \]

To solve for \( m \) and \( n \), we can add equations (1) and (2): \[ (m+n) + (m-n) = 30 + 3 \implies 2m = 33 \implies m = 16.5 \]

Now substitute \( m \) back into equation (1) to find \( n \): \[ 16.5 + n = 30 \implies n = 30 - 16.5 = 13.5 \]

Thus, the powers \( m \) and \( n \) are: \[ m = 16.5 \quad \text{and} \quad n = 13.5 \]

To verify, we calculate the product and quotient:

  1. Product: \[ a^m \cdot a^n = a^{16.5} \cdot a^{13.5} = a^{(16.5 + 13.5)} = a^{30} \] Assuming \( a = 2 \): \[ 2^{30} = 8^{10} \] (verified)

  2. Quotient: \[ \frac{a^m}{a^n} = \frac{a^{16.5}}{a^{13.5}} = a^{(16.5 - 13.5)} = a^{3} \] Assuming \( a = 2 \): \[ 2^{3} = 8 \] (verified)

Thus, the two powers that satisfy the conditions are \( m = 16.5 \) and \( n = 13.5 \).