Question

Solve each exponential equation.
63^x = 36^x+1

Answer 1

x = 6.404

Explanation:

The given equation is:

63ˣ = 36⁽ˣ⁺¹⁾

Taking log on both side

x log(63) = (x + 1) log(36)

⇒ x log(63) = x log(36) + log(36)

⇒ x log(63) - x log(36) = log(36)

⇒ x {log(63) - log(36)} = log(36)

⇒
`x = (\log(36))/(\log(63) - \log(36))`

⇒
`x = (1.556)/(1.799 - 1.556)` {∵ log(36) = 1.556 and log(63) = 1.779}

⇒ x = 6.404

Answer 2

Answer:x=6.403

Explanation:

Given

`63^x=36^(x+1)`

taking natural log both side

`\ln \left ( 63^x\right )=\ln \left ( 36\right )^(x+1)`

`x\ln \left ( 63\right )=\left ( x+1\right )\ln \left ( 36\right )`

`x\left ( \ln \left ( 63\right )-\ln \left ( 36\right )\right )=\ln \left ( 36\right )`

`x=(\ln \left ( 36\right ))/(\ln \left ( 63\right )-\ln \left ( 36\right ))`

`x=6.403`