CameraIcon

SearchIcon

MyQuestionIcon

3

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

Product Law

If 2x+4-2x+...

Question

If $2^{x + 4} - 2^{x + 2} = 3$ , then $x$ is equal to

A

0

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

B

2

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

C

- 1

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

D

- 2

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

Open in App

Solution

The correct option is D $- 2$
$2^{x + 4} - 2^{x + 2} = 3 \Rightarrow 2^{4} {.2}^{x} - 2^{2} {.2}^{x} = 3$
$\Rightarrow {16.2}^{x} - {4.2}^{x} = 3 \Rightarrow 2^{x} (16 - 4) = 3$
$2^{x} = \frac{3}{12} \Rightarrow 2^{x} = \frac{1}{4} \Rightarrow 2^{x} = 2^{- 2} \Rightarrow x = - 2$

Suggest Corrections

thumbs-up

0

Similar questions

I = \int_{2}^{3} \frac{2 x^{5} + x^{4} - 2 x^{3} + 2 x^{2} + 1}{(x^{2} + 1) (x^{4} - 1)} d x

I

x^{4} + 2 x^{3} - 4 x^{2} - 4 x + 4 = 0

2 s_{1} - s_{2} + s_{3} - s_{4}

(2 + \sqrt{3})^{x^{2} - 2 x + 1} + (2 - \sqrt{3})^{x^{2} - 2 x - 1} = \frac{2}{2 - \sqrt{3}}

, then x is equal to:

4^{2 s i n^{2} x} {.16}^{t a n^{2} x} {.2}^{4 c o s^{2} x} = 256

0 < x < \frac{π}{2}

x

is equal to ___________.

4^{x} + 2^{2 x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}

, then x is equal to

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Laws of Indices

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon