If logk x-log5 k-logx 5- k - 1- k-0- then x is equal to

Log_k x.log_5 k=log_x 5⇒ log_5 x=log_x 5 ⇒ log_x 5 =1/log_x 5⇒ (log_x 5)^2=1 ⇒ log_x 5= ± 1 ⇒ x^± 1=5 ⇒ x =5, 1/5. ...

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Fundamental Laws of Logarithms

If logk x.log...

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If $l o g_{k} x . l o g_{5} k = l o g_{x} 5, k \neq 1, k > 0$ , then x is equal to

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\frac{1}{5}

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None of these

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{(1 + x + x^{2} + x^{3})}^{5} = 15 \sum k = 0 a_{k} x^{k}

7 \sum k = 0 a_{2 k}

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