Logarithmic Function MCQ Quiz in मल्याळम - Objective Question with Answer for Logarithmic Function - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Differentiating both the sides 

\(\Rightarrow \dfrac { dy }{ dx } =-\dfrac { 99x }{ 101y }\)

Calculation

y = logn x

\(\frac{dy}{dx} = \frac{1 }{(x log^{n-1} x log^{n-2 }x ... log x)}\)

x log x log2 x log3 x ..... logn-1 x logn x \(\frac{dy}{dx}\) = logn x

Hence option 4 is correct 

Answer : 1

Solution :

log(x + y) = 2xy

Differentiating w.r.t. x, we get

\(\frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y\)

\(\frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y\)

\(\left(\frac{1}{x+y}-2 x\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y-\frac{1}{x+y}\)

\(\frac{\mathrm{d} y}{\mathrm{~d} x}\left(\frac{1}{x+y}-2 x\right)=2 y-\frac{1}{x+y}\)

\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left(2 y-\frac{1}{x+y}\right)}{\left(\frac{1}{x+y}-2 x\right)}\)

For x = 0, log(y) = 0

⇒ y = 1

\(\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{(0,1)}=\frac{\left(2-\frac{1}{0+1}\right)}{\left(\frac{1}{0+1}-0\right)}=1\)

Given the values:

log 2 = 0.2614

log 3 = 0.3521

Formula Used:

Using the logarithm property:

Calculation:

We can write as 6 = 2 x 3

Using the property:

log 6 = log 2 + log 3

Substituting the given values:

⇒ log 6 = 0.6135

Hence the value log 6 is 0.6135.

Therefore, the correct option is (2)