WAEC SSCE - Further Mathematics - 2006 (Objective)

Question 1 Report

Evaluate $\log_{0.25} 8$

$\frac{3}{2}$

$\frac{2}{3}$

$-\frac{2}{3}$

$-\frac{3}{2}$

Answer Details

We can use the change of base formula to evaluate $\log_{0.25} 8$. The change of base formula states that for any positive numbers $a$, $b$, and $c$ with $a \neq 1$, we have: $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ In this case, we can choose any base we like, but it is convenient to use base 2 because 8 and 0.25 are powers of 2. So, we can rewrite the expression as: $\log_{0.25} 8 = \frac{\log_{2} 8}{\log_{2} 0.25}$ Now, we can evaluate the logarithms on the right-hand side using the rules of logarithms. We know that $2^{3} = 8$ and $2^{-2} = 0.25$, so we have: $\log_{2} 8 = 3$ and $\log_{2} 0.25 = -2$ Substituting these values into the equation above, we get: $\log_{0.25} 8 = \frac{3}{-2} = -\frac{3}{2}$ Therefore, the value of $\log_{0.25} 8$ is $-\frac{3}{2}$. Hence, the answer is $-\frac{3}{2}$.

Read lesson note on Indices And Logarithmic Functions (WAEC)






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