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 9: Indices - Exercise 9.1

पाठ 9 Indices
Exercise 9.1 | Q 9.01

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

Solve:

2^2x + 2^x+2 − 4 × 2³ = 0

Prove that : `((x^a)/(x^b))^( a + b - c ) (( x^b)/(x^c))^( b + c - a )((x^c)/(x^a))^( c + a - b)`

Prove that :
`[ x^(a(b - c))]/[x^b(a - c)] ÷ ((x^b)/(x^a))^c = 1`

Solve : 3^x-1× 5^2y-3 = 225.

Evaluate the following:

`(12^2 xx 75^-2 xx 35 xx 400)/(48^2 xx 15^-3 xx 525)`

Evaluate the following:

`9^(5/2) - 3 xx 5^0 - (1/81)^((-1)/2)`

Solve for x:
2^2x^{- 1} − 9 x 2^{x − 2} + 1 = 0

If `root(x)("a") = root(y)("b") = root(z)("c")` and abc = 1, prove that x + y + z = 0

Prove the following:

`(x^("p"("q"-"r")))/(x^("q"("p"-"r"))) ÷ (x^"q"/x^"p")^"r"` = 1