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Logarithm Properties PYQ | JEE Main Mathematics 2025

Solve this JEE Main 2025 April Mathematics PYQ on logarithms using the change of base formula, power rule, and inverse logarithm properties. Learn the

 

❓ Question

Find the value of

811log53+27log936+34log79.81^{\frac{1}{\log_5 3}} +27^{\log_9 36} +3^{\frac{4}{\log_7 9}}.

Logarithm Properties PYQ | JEE Main Mathematics 2025

✍️ Solution

First term

811log53=(34)1log53=34log53.81^{\frac{1}{\log_5 3}} =(3^4)^{\frac{1}{\log_5 3}} =3^{\frac{4}{\log_5 3}}.

Using

1logba=logab,\frac{1}{\log_b a}=\log_a b,

we get

34log35.3^{4\log_3 5}.

Using

aklogab=bk,a^{k\log_a b}=b^k,
34log35=54=625.3^{4\log_3 5}=5^4=625.

Second term

27log936=(33)log936=33log936.27^{\log_9 36} =(3^3)^{\log_9 36} =3^{3\log_9 36}.

Now,

36=62,9=32.36=6^2,\qquad 9=3^2.

Hence,

3log936=3log336log39=32log362=3log36.3\log_9 36 = 3\cdot\frac{\log_3 36}{\log_3 9} = 3\cdot\frac{2\log_3 6}{2} = 3\log_3 6.

Therefore,

33log36=63=216.3^{3\log_3 6}=6^3=216.

Third term

34log79=34log97.3^{\frac{4}{\log_7 9}} = 3^{4\log_9 7}.

Since

9=32,9=3^2,
log97=log372.\log_9 7=\frac{\log_3 7}{2}.

Thus,

34log97=32log37=72=49.3^{4\log_9 7} = 3^{2\log_3 7} = 7^2 = 49.

Adding all three terms,

625+216+49=890.
625+216+49=890.

Logarithm Properties PYQ | JEE Main Mathematics 2025

✅ Final Answer

890\boxed{890}

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