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

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

shaalaa.com

Solving Exponential Equations

Is there an error in this question or solution?

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

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

Solve for x : (49)^{x + 4} = 7²x (343)^{x + 1}

Solve for x: `4^(x-1) × (0.5)^(3 - 2x) = (1/8)^-x`

If a^x = b^y = c^z and b² = ac, prove that: y = `[2xz]/[x + z]`

Solve x and y if : ( √32 )^x ÷ 2^{y + 1} = 1 and 8^y - 16^{4 - x/2} = 0

Solve for x:
1 = p^x

Solve for x:

`"p"^-5 = (1)/"p"^(x + 1)`

Find the value of (8p)^p if 9^{p + 2} - 9^p = 240.

If 2^x = 3^y = 12^z ; show that `(1)/z = (1)/y + (2)/x`.

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

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

Prove the following:

`sqrt(x^-1 y) · sqrt(y^-1 z) · sqrt(z^-1 x)` = 1