Question

If $log_{10}5 = a$ and $log_{10}3 = b$ then :

• A. $log_{30} 8 = \frac{3(1-a)}{b+1}$
• B. $log_{40} 15 = \frac{a+b}{3-2a}$
• C. $log_{243} 32 = \frac{1-a}{b}$
• D. $log_{5} 3 = \frac{b}{a}$

Answer 1

Answer: (A), (B), (C), (D)

(A), (B), (C), (D)

Answer 2

Given $log_{10}5 = a$ and $log_{10}3 = b$.

We know that $log_{10}10 = 1$.

Using the property $log_b (xy) = log_b x + log_b y$, we have $log_{10}10 = log_{10}(5 \times 2) = log_{10}5 + log_{10}2$.

So, $1 = a + log_{10}2$, which gives $log_{10}2 = 1 - a$.

Now, we evaluate each option using the change of base formula $log_c x = \frac{log_d x}{log_d c}$ with $d=10$.

(A) $log_{30} 8$: $log_{30} 8 = \frac{log_{10} 8}{log_{10} 30} = \frac{log_{10} (2^3)}{log_{10} (3 \times 10)} = \frac{3 log_{10} 2}{log_{10} 3 + log_{10} 10} = \frac{3(1-a)}{b + 1}$. Option (A) is correct.

(B) $log_{40} 15$: $log_{40} 15 = \frac{log_{10} 15}{log_{10} 40} = \frac{log_{10} (3 \times 5)}{log_{10} (4 \times 10)} = \frac{log_{10} 3 + log_{10} 5}{log_{10} (2^2 \times 10)} = \frac{log_{10} 3 + log_{10} 5}{log_{10} (2^2) + log_{10} 10} = \frac{b + a}{2 log_{10} 2 + 1}$.

Substitute $log_{10} 2 = 1-a$: $\frac{a+b}{2(1-a) + 1} = \frac{a+b}{2 - 2a + 1} = \frac{a+b}{3 - 2a}$. Option (B) is correct.

(C) $log_{243} 32$: $log_{243} 32 = \frac{log_{10} 32}{log_{10} 243} = \frac{log_{10} (2^5)}{log_{10} (3^5)} = \frac{5 log_{10} 2}{5 log_{10} 3} = \frac{log_{10} 2}{log_{10} 3}$.

Substitute $log_{10} 2 = 1-a$ and $log_{10} 3 = b$: $\frac{1-a}{b}$. Option (C) is correct.

(D) $log_{5} 3$: $log_{5} 3 = \frac{log_{10} 3}{log_{10} 5} = \frac{b}{a}$. Option (D) is correct.

All four options are correct.