Question

Write the following logarithmic form:
i) 8³ = 512
ii) 7⁻² = 1/49
iii) 10⁻² = 0.01

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

(i) $log_{8}512=3$

(ii) $log_{7}\frac{1}{49}=-2$

(iii) $log_{10}0.01=-2$

Step-by-step explanation:

In the question,

We have to write the following equations in the logarithms term,

So,

As, we know that from the Properties of Logarithms that,

$loga^{b}=bloga$

(i) 8³ = 512

Now,

Taking log to the base 8 on the both sides we get,

$8^{3}=512\\log_{8}(8)^{3}=log_{8}512\\3log_{8}8=log_{8}512\\So,\\log_{8}512=3$

(ii) 7⁻² = 1/49

Now,

On taking log to the base 7 on both sides we get,

$7^{-2}=\frac{1}{49}\\log_{7}(7^{-2})=log_{7}\frac{1}{49}\\-2log_{7}7=log_{7}\frac{1}{49}\\\\So,\\log_{7}\frac{1}{49}=-2$

(iii) 10⁻² = 0.01

Now,

On taking log to the base 10 on the both sides we get,

$10^{-2}=0.01\\log_{10}(10)^{-2}=log_{10}0.01\\-2.log_{10}10=log_{10}0.01\\So,\\log_{10}0.01=-2$