Solve 2^(x-1) = 8 ?

2^(x-1)=8
2^(x-1)=2^3

as the bases are same we can equate the powers, so we get
x-1=3
x=3+1
x=4

x - 1 = 3
x = 4

it relatively is fairly user-friendly question on left hand component capacity of two is 2x+one million so on suitable hand component write 8 in capacity of two that's 2^3 now on the two factors we've with their powers so thier powers are equivalent to a minimum of one yet another like 2x+one million = 3 2x = 3-one million 2x = 2 x = 2/2 x = one million examine it by utilising putting x=one million on your equation

2^(x - 1) = 8
2^(x - 1) = 2^3
x - 1 = 3
x = 3 + 1
x = 4

Since 2^3 = 8, 2^3 = 2^(x - 1)

By comparison, 3 = x - 1 and x = 4.

2^(x-1) = 8

8 = 2^3

2^(x-1) = 2^3

x-1 = 3

x=4

4

or take logs:
(x-1) log2=log8
divide by log2
x-1=log8/log2=3
add the 1
x=4

2^(x - 1) = 8
2^(x - 1) = 2^3
Therefore,
x - 1 = 3
x = 4 (ans)

2^(x-1) = 8

2^(x-1) = 2^3

(x-1) = 3
x = 4