Solve: 3log₄(9) + 8log₄(1/3) Logarithmic Expression

Because the coefficients represent powers, not simple addition! You must first use the power rule: $n\log_a x = \log_a x^n$. Only after converting can you use the addition rule.

Use the negative exponent rule: $\log_a x^{-n} = -n\log_a x$. So $\log_4 9^{-1} = -\log_4 9$. The negative sign comes outside the logarithm!

When you raise $\frac{1}{3}$ to the 8th power, you get $\frac{1^8}{3^8} = \frac{1}{3^8} = \frac{1}{6561}$. This is a much smaller number than the original fraction!

Use the rule: $\log_a x + \log_a y = \log_a(xy)$. So multiply the arguments: $729 \times \frac{1}{6561} = \frac{729}{6561} = \frac{1}{9}$.

Because $729 = 9^3$ and $6561 = 3^8 = (3^4)^2 = 81^2$. Actually, $6561 = 9^4$, so $\frac{729}{6561} = \frac{9^3}{9^4} = 9^{-1} = \frac{1}{9}$!

Yes! Calculate $3\log_4 9 + 8\log_4 \frac{1}{3}$ and $-\log_4 9$ separately. They should give the same decimal value (approximately -1.585).