The antilog of 2-5 to the base 243 is

Name: JEE- Grade 11- Mathematics- Fundamentals of Mathematics 13- Session 13- D09
Uploaded: 2022-07-04T10:36:42+05:30
Description: Antilog of 25 to the base 243 is nbsp; (243)^2/5=(3^5)^2/5=3^2=9 ...

Antilog of 25 to the base 243 is nbsp; (243)^2/5=(3^5)^2/5=3^2=9 ...

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Standard XII

Mathematics

Fundamental Laws of Logarithms

The antilog o...

Question

The antilog of $\frac{2}{5}$ to the base $243$ is

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Solution

Antilog of $\frac{2}{5}$ to the base $243$ is
$(243)^{2 / 5} = (3^{5})^{2 / 5} = 3^{2} = 9$

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Fundamental Laws of Logarithms

Standard XII Mathematics

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The antilog of 25 to the base 243 is

Antilog of 25 to the base 243 is (243)2/5=(35)2/5=32=9

The antilog of $\frac{2}{5}$ to the base $243$ is

Antilog of $\frac{2}{5}$ to the base $243$ is
$(243)^{2 / 5} = (3^{5})^{2 / 5} = 3^{2} = 9$