Solve for X : 3(2x + 1) - 2x + 2 + 5 = 0

Solve for x : 3(2^x + 1) - 2^{x + 2} + 5 = 0

Sum

3(2^x + 1) - 2^{x + 2} + 5 = 0
⇒ 3 x 2^x + 3 - 2^x x 2² + 5 = 0
⇒ 2^x ( 3 - 4 ) + 8 = 0
⇒ - 2^x= - 8
⇒ 2^x= 2³⇒ x = 3

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Solving Exponential Equations

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Chapter 7: Indices [Exponents] - Exercise 7 (B) [Page 101]

Chapter 7 Indices [Exponents]
Exercise 7 (B) | Q 14.5 | Page 101

Solve for x : 2^2x+1 = 8

Solve for x:

`3^(4x + 1) = (27)^(x + 1)`

Solve : 8 x 2^2x + 4 x 2^{x + 1} = 1 + 2^x

Solve:

2^2x + 2^x+2 − 4 × 2³ = 0

Find the values of m and n if :
`4^(2m) = ( root(3)(16))^(-6/n) = (sqrt8)^2`

Solve for x:
2^2x + 2^{x +2} - 4 x 2³ = 0

Find the value of k in each of the following:

`(sqrt(9))^-7 xx (sqrt(3))^-5` = 3^k

If a = `2^(1/3) - 2^((-1)/3)`, prove that 2a³ + 6a = 3

If 2400 = 2^x x 3^y x 5^z, find the numerical value of x, y, z. Find the value of 2^-x x 3^y x 5^z as fraction.

Find the value of 'a' and 'b' if:

`(sqrt243)^"a" ÷ 3^("b" + 1)` = 1 and `27^"b" - 81^(4 -"a"/2)` = 0