Solve the following equations for x:

Question:

Solve the following equations for x: 

(i) $7^{2 x+3}=1$

(ii) $2^{x+1}=4^{x-3}$

(iii) $2^{5 x+3}=8^{x+3}$

(iv) $4^{2 x}=1 / 32$

(v) $4^{x-1} \times(0.5)^{3-2 x}=(1 / 8)^{x}$

(vi) $2^{3 x-7}=256$

Solution:

(i) We have,

$\Rightarrow 7^{2 x+3}=1$

$\Rightarrow 7^{2 x+3}=7^{0}$

$\Rightarrow 2 x+3=0$

$\Rightarrow 2 x=-3$

$\Rightarrow x=-3 / 2$

(ii) We have,

$=2^{x+1}=4^{x-3}$

$=2^{x+1}=2^{2 x-6}$

$=x+1=2 x-6$

$=x=7$

(iii) We have,

$=2^{5 x+3}=8^{x+3}$

$=2^{5 x+3}=2^{3 x+9}$

$=5 x+3=3 x+9$

$=2 x=6$

$=x=3$

(iv) We have,

$=4^{2 x}=1 / 32$

$=2^{4 x}=1 / 2^{5}$

$=2^{4 x}=2^{-5}$

$=4 x=-5$

$x=-5 / 4$

(v) We have,

$4^{x-1} \times(0.5)^{3-2 x}=(1 / 8)^{x}$

$2^{2 x-2} \times(1 / 2)^{3-2 x}=(1 / 2)^{3 x}$

$2^{2 x-2} \times 2^{2 x-3}=(1 / 2)^{3 x}$

$2^{2 x-2+2 x-3}=(1 / 2)^{3 x}$

$2^{4 x-5}=2^{-3 x}$

$4 x-5=-3 x$

$7 x=5$

$x=5 / 7$

(vi) $2^{3 x-7}=256$

$2^{3 x-7}=2^{8}$

$3 x-7=8$

$3 x=15$

$x=5$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now