Solve, 4(2x−1)−2(x−5)=5(x+1)+3
Given, 4(2x−1)−2(x−5)=5(x+1)+3
⇒(8x-4)-(2x-10)=5x+5+3
⇒8x-4-2x+10=5x+8
⇒6x-6=5x+8
⇒6x-5x=8+6
∴x=14
Hence, the required value of x=14.
Find x, if:
(i) 42x=132
(ii) √2x+3=16
(iii) (√35)x+1=12527
(iv) (3√23)x−1=278
Solve the following equations for x:
(i) 72x+3=1
(ii) 2x+1=4x−3
(iii)25x+3=8x+3
(iv) 42x=132
(v)4x−1×(0.5)3−2x=(18)x
(vi)23x−7=256
Find the values of x in each of the following:
(i)25x÷2x=5√220
(ii)(23)4=(22)x
(iii)(35)x(53)2x=12527
(iv)5x−2×32x−3=135
(v)2x−7×5x−4=1250
(vi)(3√4)2x+12=132
(vii)52x+3=1
(viii) (13)√x=44−34−6
(ix)(√35)x+1=12527