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

Solve for X: 1 = Px | Shaalaa.com

Solve for x:
1 = p^x

योग

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

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Solving Exponential Equations

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अध्याय 6: Indices - Exercise 9.1

अध्याय 6 Indices
Exercise 9.1 | Q 9.06

Find x, if : `( sqrt(3/5))^( x + 1) = 125/27`

Solve for x : 9^x+2= 720 + 9^x

If 5^{x + 1} = 25^{x - 2}, find the value of 3^{x - 3} × 2^{3 - x}.

If m ≠ n and (m + n)^-1 (m^-1 + n^-1) = m^xn^y, show that : x + y + 2 = 0

Evaluate : `[(-2/3)^-2]^3 xx (1/3)^-4 xx 3^-1 xx 1/6`

Solve : 3^x-1× 5^2y-3 = 225.

Evaluate the following:

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

Solve for x:
2^2x+3 - 9 x 2^x+ 1 = 0

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

Prove the following:

`(x^("a"+"b")/x^"c")^("a"-"b") · (x^("c"+"a")/(x^"b"))^("c"-"a") · ((x^("b"+"c"))/(x"a"))^("b"-"c")` = 1

Solve for X: 1 = Px