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

Solve for X: 1 = Px | Shaalaa.com

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 6: Indices - Exercise 9.1

पाठ 6 Indices
Exercise 9.1 | Q 9.06

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

Solve for x : 3(2^x + 1) - 2^{x + 2} + 5 = 0

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

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

Evaluate the following:

`(2^6 xx 5^-4 xx 3^-3 xx 4^2)/(8^3 xx 15^-3 xx 25^-1)`

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

Find the value of k in each of the following:

`(1/3)^-4 ÷ 9^((-1)/(3)` = 3^k

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

Prove the following:

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

Prove the following:
(x^a)^b-c x (x^b)^c-a x (x^c)^a-b= 1

Solve for X: 1 = Px