Let f(a)=\log_{a^{2}}25 \: \: and \: \: g(a)=\log_{a}5 then f(a)=g(a) holds for a belonging to

  • Option 1)

    R

  • Option 2)

    (0,1)\cup (1,+\infty )

  • Option 3)

    \phi

  • Option 4)

    None of these

 

Answers (1)

As we learnt in 

Logarithmic Function -

f\left ( x \right )= \log\: _{a}x

base:a> 0,a\neq 1

domin:x> 0

Range:f\left ( x \right )\in R

- wherein

 

\\f(a)=log_{a^{2}}^{25};\ \ g(a)=log_{a}^{5}\\*\\*=log_{a^{2}}^{5^{2}}\\*\\=\frac{2}{2}\: log_{a}^{5}=log_{a}^{5}\\*\\* But\: \: a\neq 1\\*\\*Thus\: \: \: a\epsilon \left ( 0,1 \right )\cup \left ( 1,\infty \right )


Option 1)

R

Incorrect

Option 2)

(0,1)\cup (1,+\infty )

Correct

Option 3)

\phi

Incorrect

Option 4)

None of these

Incorrect

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