Solve for X: P − 5 = 1 P X + 1 - Mathematics

Solve for x:

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

बेरीज

`"p"^-5 = (1)/"p"^(x + 1)`
⇒ p^-5 x p^{x + 1} = 1
⇒ `"p"^(-5 + x +1)` = 1
⇒ p^x-4= p⁰
⇒ x - 4 = 0
⇒ x = 4.

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Solving Exponential Equations

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पाठ 9: Indices - Exercise 9.1

पाठ 9 Indices
Exercise 9.1 | Q 9.08

Solve for x:

`3^(4x + 1) = (27)^(x + 1)`

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

Solve : 2^2x + 2^x+2 - 4 x 2³ = 0

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

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

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

Solve for x:
9^x+4= 3² x (27)^x+1

If a^x = b^y = c^z and abc = 1, show that

`(1)/x + (1)/y + (1)/z` = 0.

Find the value of 'a' and 'b' if:

`(sqrt243)^"a" ÷ 3^("b" + 1)` = 1 and `27^"b" - 81^(4 -"a"/2)` = 0

Prove the following:

`("a"^"m"/"a"^"n")^("m"+"n"+1) ·("a"^"n"/"a"^1)^("n" + 1-"m").("a"^1/"a"^"m")^(1+"m"-"n")`