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

Solve for x:

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

Sum

`"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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Chapter 9: Indices - Exercise 9.1

Chapter 9 Indices
Exercise 9.1 | Q 9.08

Solve : `[3^x]^2` : 3x = 9 : 1

Solve : 2^2x + 2^x+2 - 4 x 2³ = 0

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

Prove that :
`[ x^(a(b - c))]/[x^b(a - c)] ÷ ((x^b)/(x^a))^c = 1`

Evaluate the following:

`9^(5/2) - 3 xx 5^0 - (1/81)^((-1)/2)`

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

Solve for x:
1 = p^x

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

Solve for x:
5^x2 : 5^x= 25 : 1

Find the value of k in each of the following:

`(1/3)^-4 ÷ 9^((-1)/(3)` = 3^k