1 = px⇒ p0 = px ......(Using a0 = 1)⇒ x = 0.

Solve for X: 1 = Px - Mathematics

Solve for x:
1 = p^x

बेरीज

1 = p^x
⇒ p⁰ = p^x ......(Using a⁰= 1)
⇒ x = 0.

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Solving Exponential Equations

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पाठ 9: Indices - Exercise 9.1

पाठ 9 Indices
Exercise 9.1 | Q 9.06

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

Solve : `(sqrt(3))^( x - 3 ) = ( root(4)(3))^( x + 1 )`

If 5^-P = 4^-q = 20^r, show that : `1/p + 1/q + 1/r = 0`

Evaluate the following:

`(2^3 xx 3^5 xx 24^2)/(12^2 xx 18^3 xx 27)`

Solve for x:
3 x 7^x = 7 x 3^x

Solve for x:
2^{x + 3} + 2^{x + 1} = 320

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

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

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

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