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Given log 2 = a, log 3 = b, express in terms of a and b. log 4.8 - Mathematics

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Question

Given log 2 = a, log 3 = b, express in terms of a and b.

log 4.8

Sum
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Solution

Given: log 2 = a, log 3 = b

To find: log 4.8

Calculation:

Step 1: Express 4.8 as a product of known quantities

`4.8 = 4 xx 1.2 = 2^2 xx (6/5)`

Thus:

`log 4.8 = log(2^2 xx 6/5)`

Step 2: Apply the logarithmic properties:

Using the properties of logarithms, we can break this down:

`log 4.8 = log(2^2) + log(6/5)`

Now, use `log  a/b = log a - log b`:

log 4.8 = 2 log 2 + log 6 – log 5

Now substitute the given values:

log 4.8 = 2a + (log 2 + log 3) – log 5

Since log 2 = a, we have:

log 4.8 = 2a + a + b – log 5

Finally, we need to express log 5 in terms of a and b.

Using log 10 = 1, we know:

log 10 = log(2 × 5) = log 2 + log 5

So:

1 = a + log 5

⇒ log 5 = 1 – a

Step 3: Final expression for log 4.8:

Now substitute log 5 = 1 – a into the equation:

log 4.8 = 3a + b – (1 – a)

Simplify:

log 4.8 = 3a + b – 1 + a

log 4.8 = 4a + b – 1

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Chapter 7: Logarithms - MISCELLANEOUS EXERCISE [Page 77]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 7 Logarithms
MISCELLANEOUS EXERCISE | Q 7. (ii) | Page 77
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