Math, asked by laxmipriyakhatoi5, 23 days ago

Prove that a^2b^3 =100 where 2 loga + 3 logb = 2​

Answers

Answered by anindyaadhikari13
4

Solution:

Given That:

→ 2 ㏒(a) + 3 ㏒(b) = 2

Can be written as:

→ ㏒(a²) + ㏒(b³) = 2

→ ㏒(a²b³) = 2

We know that:

→ ㏒(m) = n or 10ⁿ = m

Similarly:

→ 10² = a²b³

→ a²b³ = 10²

→ a²b³ = 100

Hence Proved...!!

To Know More:

Laws of Logarithms.

 \rm 1. \:  \:  {a}^{n} = b \implies log_{a}(b)  = n

 \rm 2. \:  \: log_{a}(1)  = 0, \: a \neq0,1

 \rm 3. \:  \: log_{a}(a)  = 1, \: a \neq0,1

 \rm 4. \:  \: log_{a}(x)  = log_{a}(y) \implies x = y

 \rm 5. \:  \: log_{e}(x) =  ln(x)

 \rm6. \:  \:  log_{a}(x) + log_{a}(y) = log_{a}(xy)

 \rm7. \:  \:  log_{a}(x) - log_{a}(y) = log_{a} \bigg( \dfrac{x}{y} \bigg)

 \rm 8. \:  \: log_{a}( {x}^{n} ) =  n\log_{a}(x)

 \rm 9. \:  \:  log_{a}(m) =  \dfrac{ log_{b}(m) }{ log_{b}(a) },m > 0,b > 0,a \ne1,b \ne1

 \rm 10. \:  \: log_{a}(b) = \dfrac{1}{ log_{b}(a) }

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