22x+1= 8⇒ 22x+1= 23⇒ 2x + 1 = 3⇒ 2x = 2⇒ x = 1.

Solve for X: 22x+1= 8

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

बेरीज

2^2x+1= 8
⇒ 2^2x+1= 2³
⇒ 2x + 1 = 3
⇒ 2x = 2
⇒ x = 1.

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Solving Exponential Equations

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पाठ 6: Indices - Exercise 9.1

पाठ 6 Indices
Exercise 9.1 | Q 9.01

Solve for x:

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

Solve : 4^{x - 2} - 2^{x + 1} = 0

If 3^{x + 1} = 9^{x - 3}, find the value of 2^{1 + x}.

Evaluate : `4/(216)^(-2/3) + 1/(256)^(-3/4) + 2/(243)^(-1/5)`

Evaluate the following:

`(27)^(2/3) xx 8^((-1)/6) ÷ 18^((-1)/2)`

Solve for x:
`9 xx 3^x = (27)^(2x - 5)`

Solve for x:

9 x 81^x = `(1)/(27^(x - 3)`

Solve for x:
5^x2 : 5^x= 25 : 1

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.

Prove the following:

`root("ab")(x^"a"/x^"b")·root("bc")(x^"b"/x^"c")·root("ca")(x^"c"/x^"a")` = 1