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

Solve for x:

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

योग

`"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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अध्याय 9: Indices - Exercise 9.1

अध्याय 9 Indices
Exercise 9.1 | Q 9.08

Solve for x : 2^2x+1 = 8

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

If a^x = b, b^y = c and c^z = a, prove that : xyz = 1.

If `((a^-1b^2 )/(a^2b^-4))^7 ÷ (( a^3b^-5)/(a^-2b^3))^-5 = a^x . b^y` , find x + y.

Evaluate the following:

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

Solve for x:
`9 xx 3^x = (27)^(2x - 5)`

Solve for x:
9^x+4= 3² x (27)^x+1

Find the value of k in each of the following:

`(sqrt(9))^-7 xx (sqrt(3))^-5` = 3^k

Find the value of k in each of the following:

`(1/3)^-4 ÷ 9^((-1)/(3)` = 3^k

If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)³ = 27xyz