If $l o g_{8} m = 3.5$ and $l o g_{2} n = 7$ , then the value of $m$ in terms of $n$ is

n \sqrt{n}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 n^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $n \sqrt{n}$
Given, $l o g_{8} m = 3.5$ and $l o g_{2} n = 7$
Therefore, $m = 8^{3.5}$ and $n = 2^{7}$
$⟹ m = 8^{\frac{35}{10}}$
$⟹ m = (2^{7})^{3 / 2}$
$⟹ m = n^{3 / 2} = n \sqrt{n}$

Suggest Corrections

Similar questions

{log}_{8} m = 3.5

{log}_{2} n = 7

If $n$ is odd, then $1 + 3 + 5 + 7 + . . . . . +$ to $n$ terms is equal to

(a) $(n^{2} + 1)$

(b) $(n^{2} - 1)$

(d) $(2 n^{2} + 1)$

lim n \to \infty [\frac{2 n}{2 n^{2} - 1} cos \frac{n + 1}{2 n - 1} - \frac{n}{1 - 2 n} \cdot \frac{n {(- 1)}^{n}}{n^{2} + 1}]

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

lim n \to \infty [\frac{2 n}{2 n^{2} - 1} cos \frac{n + 1}{2 n - 1} - \frac{n}{1 - 2 n} . \frac{n (- 1)^{n}}{n^{2} + 1}]

Join BYJU'S Learning Program