Question

If log16 X + log4 x + log2 X = 14, then x =
(A) 16.
(B) 32
(C) 64
(D) None of these​

Answer 1

Answer:

D)NONE OF THESE

Step-by-step explanation:

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Answer 2

Answer:

(D) None of these

Step-by-step explanation:

Given:- $log_{16} x + log_{4} x + log_{2} x = 14$

To Find:- Value of x.

Solution:-

Acc. to the question, $log_{16} x + log_{4} x + log_{2} x = 14$

As we know, acc. to logarithmic formula,

                   $log_{p} q = \frac{log q}{log p}$

∴ $\frac{log x}{log 16} + \frac{log x}{log 4} + \frac{log x}{log 2} = 14$

⇒ $log x (\frac{1}{log 16} + \frac{1}{log 4} + \frac{1}{log 2} ) = 14$

⇒ $log x (\frac{1}{log 2^{4} } + \frac{1}{log 2^{2} } + \frac{1}{log 2} ) = 14$    

⇒ $log x (\frac{1}{4 log 2 } + \frac{1}{2log 2 } + \frac{1}{log 2} ) = 14$           [∵ $logx^{n} = n log x$]          

⇒ $log x (\frac{1 + 2 + 4}{4log 2} ) = 14$

⇒ $log x (\frac{7}{4log 2} ) = 14$

⇒ $log x = \frac{14 * 4log2}{7}$

⇒ $log x = 8 log 2$

⇒ $log x = log 2^{8}$          [∵ $n log x = logx^{n}$]

⇒ $x = 2^{8}$

⇒ $x = 256.$

Hence, option (D) is the correct option.

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