If logx 625 - -4 then x - 10 is

The correct option is D 2 #160;625 = x^ 4 ⇒ 5 ^ 4 = 1 x ^ 4 So x = 1 5 x×10 = 2 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Integration of Piecewise Continuous Functions

If logx 625...

Question

If ${log}_{x} 625 = - 4$ then $x \times 10$ is

0

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

- 2

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

- 1

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

2

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

Open in App

Solution

Suggest Corrections

Similar questions

{log}_{x} 625 = - 4

x

{(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{10} x^{10}

{(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2}

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

If $(2 x^{2} - x - 1)^{5} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . . . . . . a_{10} x^{10}$ then $a_{2} + a_{4} + a_{6} + a_{8} + a_{10}$ =

x = 2 + i

\frac{1}{0! 8!} + \frac{1}{2! 6!} + \frac{1}{2! 4! 4!} = \frac{2^{a - 1}}{m!}

x^{3} - 5 x^{2} + 9 x + 10

Join BYJU'S Learning Program