Question

E-4. (log₂10) . (log₂80) - (log₂5) . (log₂160) is equal to :

• A. log₂5
• B. log₂20
• C. log₂10

Answer 1

The expression evaluates to 4, which is equivalent to log₂16. None of the provided options match this result.

Answer 2

Let the given expression be $E$. $E = (\log₂10) \cdot (\log₂80) - (\log₂5) \cdot (\log₂160)$

We use the properties of logarithms: $\log_b(xy) = \log_b x + \log_b y$ and $\log_b b^n = n$. Let's express the numbers inside the logarithms in terms of powers of 2 and 5. $10 = 2 \times 5$ $80 = 16 \times 5 = 2^4 \times 5$ $160 = 32 \times 5 = 2^5 \times 5$

Now, let's express each logarithmic term: $\log₂10 = \log₂(2 \times 5) = \log₂2 + \log₂5 = 1 + \log₂5$ $\log₂80 = \log₂(2^4 \times 5) = \log₂2^4 + \log₂5 = 4 \log₂2 + \log₂5 = 4 + \log₂5$ $\log₂5 = \log₂5$ $\log₂160 = \log₂(2^5 \times 5) = \log₂2^5 + \log₂5 = 5 \log₂2 + \log₂5 = 5 + \log₂5$

Let $x = \log₂5$. Substitute these expressions into the given expression $E$: $E = (1 + x)(4 + x) - (x)(5 + x)$

Now, expand and simplify the expression: $E = (1 \times 4 + 1 \times x + x \times 4 + x \times x) - (x \times 5 + x \times x)$ $E = (4 + x + 4x + x^2) - (5x + x^2)$ $E = (4 + 5x + x^2) - (5x + x^2)$ $E = 4 + 5x + x^2 - 5x - x^2$ $E = 4$

The value of the expression is 4. This is equivalent to $\log₂16$.