Question

If log2 x + log4 x + log16 x = 7/2 find the value of x​

Answer 1

GiveN :-

$\sf \: log(2x) + log(4x) + log(16x) = \frac{7}{2}$

To FinD :-

• The value of x

SolutioN :-

$\sf \: log(2x) + log(4x) + log(16x) = \frac{7}{2} \\ \\ \sf \: log(2x) + log(2 \times 2x) + log(8 \times 2x) = \frac{7}{2}$

Using the identity →

• $\sf log(m \times n) = log(m) + log(n)$

$\sf log(2x) + log(2) + log(2x) + log(8) + log(2x) = \frac{7}{2} \\ \\ \sf \: 3 log(2x) + log(8) + log(2) = \frac{7}{2} \\ \\ \sf \: 3 log(2x) + log(2 \times 2 \times 2) + log(2) = \frac{7}{2} \\ \\ \sf \: 3 log(2x) + log( {2}^{3} ) + log(2) = \frac{7}{2}$

Using the identity :-

• $\sf \: log( {m}^{n} ) = n \: log(m)$

$\sf3 log(2x) + 3 log(2) + log(2) = \frac{7}{2} \\ \\ \sf \: 3 log(2x) + 4 log(2) = \frac{7}{2}$

We know that :-

• $\sf \: log(2) = 0.3$

$\sf3 log(2x) + 4 \times 0.3 = \frac{7}{2} \\ \\ \sf \: 3 log(2x) + 1.2 = 3.5 \\ \\ \sf \: 3 log(2x) = 3.5 - 1.2 \\ \\ \sf3 log(2x) = 2.3 \\ \\ \sf log(2x) = \frac{2.3}{3} \\ \\ \sf \: log(2x) = 0.76$

We know that :-

• $\sf \: 0.76 = log(5.8)$

$\sf log(2x) = log(5.8) \\ \\ \sf \: 2x = 5.8 \\ \\ \sf \: x = \frac{5.8}{2} \\ \\ </strong><strong> </strong><strong>\</strong><strong>h</strong><strong>u</strong><strong>g</strong><strong>e</strong><strong>{</strong><strong> \boxed{ \sf \: x = 2.9</strong><strong>}</strong><strong>}$