Correct Answer - Option 2 : 622
Concept:
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\(a^{log_{b}a}=\sqrt{a}\), If b = a2, a > 0, b > 0, b \( \neq\) 1.
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\(a^{log_{b}a}=a^2\), If a = b2, a > 0, b > 0, b \( \neq\) 1
Calculation:
Given: \(25^{log_{5}25}-9^{log_{81}9}=7^{log_{7}x}\)
By using the properties mentioned in the concept part we get
⇒ \(25^2-\sqrt9=7^{log_{7}x}\)
⇒ \(625-3=7^{log_{7}x}\)
⇒ \(622=7^{log_{7}x}\)
By taking log on both sides to the base 7 we get
⇒ \({log_{7}622}={log_{7}x} \times {log_{7}7}\)
⇒ \({log_{7}622}={log_{7}x}\)
By comparing LHS and RHS, we get x = 622
Hence, option 2 is correct.