Question

81^(1/log5 (3))+27^(log9(36))+3^(4/log7(9))=​

Answer 1

Required to Know

• Log Rules

$\rightarrow \log_{a}b=\dfrac{1}{\log_{b}a}$ …[1]

$\rightarrow \log_{a^{m}}b^{n}=\dfrac{n}{m} \log_{a}{b}$ …[2]

$\rightarrow a^{\log_{a}b}=b$ …[3]

• Exponent Rules

$\rightarrow (a^{m})^{n}=a^{mn}$ …[1]

Solution

Given: $81^{(\dfrac{1}{\log_{5}3} )}+27^{(\log_{9}36)}+3^{(\dfrac{4}{\log_{7}9} )}$

Using log rule [1],

$=81^{(\log_{3}5 )}+27^{(\log_{9}36)}+3^{(4\log_{9}7)}$

Using exponent rule [1],

$=(3)^{4(\log_{3}5 )}+3^{3(\log_{3^2}6^2)}+3^{(4\log_{3^{2}}7)}$

Using log rule [2],

$=(3)^{4(\log_{3}5 )}+3^{3(\frac{2}{2} \log_{3}6)}+3^{(\frac{4}{2} \log_{3}7)}$

Using log rule [3],

$=5^{4}+6^{3}+7^{2}$

$=625+216+49=\boxed{890}$

Hence the required answer is $\boxed{890}$.