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

Find x, if : 4^2x = `1/32`

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

Solve for x : (a^{3x + 5})². (a^x)⁴ = a^{8x + 12}

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

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

Solve for x:

`"p"^-5 = (1)/"p"^(x + 1)`

Find the value of k in each of the following:

`root(4)root(3)(x^2)` = x^k

If a^x = b^y = c^z and b² = ac, prove that y = `(2xz)/(z + x)`

Prove the following:

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