Question

logarithms question ​

Answer 1

Answer

8

Solution

log81 - log3 = log(a)

We can write it as

log3⁴ - log3 = log(a)

This becomes

4log3 - log3 = log(a)

Using $\sf log_a(m^p) = plog_a(m)$

This gives

(4 - 1)log3 = loga

→ 3log3 = loga

→ log3³ = loga

Using, $\sf plog_am = log_a(m^p)$

→ a = 3³

→ a = 27

So,

$\sf 4^{log_9a}$

$\sf\implies 4^{log_927}$

$\sf\implies 4^{log_{3^2}(3^3)}$

This becomes

$\sf\implies 4^{\frac{3}{2}log_3(3)}$

Using, $\sf log_{a^q}m = \frac{1}{q}log_am$

$\sf\implies 4^{\frac{3}{2} (1)}$

$\sf\implies 4^{\frac{3}{2}} = 8$

Answer 2

Answer:

8.

Step-by-step explanation:

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