Question

What is the solution of the equation 4^(x + 1) = 21? Round your answer to the nearest ten-thousandth.

Answer 1

Answer:
`x`≈
`1.196`

Explanation:

Given the equation
`4^((x + 1)) = 21` you need to solve for the variable "x".

Remember that according to the logarithm properties:

`log_b(b)=1`

`log(a)^n=nlog(a)`

Then, you can apply
`log_4` on both sides of the equation:

`log_4(4)^((x + 1)) = log_4(21)\\\\(x + 1)log_4(4) = log_4(21)\\\(x + 1) = log_4(21)`

Apply the Change of base formula:

`log_b(x) = (log_a( x))/(log_a(b))`

Then you get:

`x =(log(21))/(log(4))-1`

`x`≈
`1.196`

Answer 2

For this case we must solve the following equation:

`4 ^ {x + 1} = 21`

We find Neperian logarithm on both sides:

`ln (4 ^ {x + 1}) = ln (21)`

According to the rules of Neperian logarithm we have:

`(x + 1) ln (4) = ln (21)`

We apply distributive property:

`xln (4) + ln (4) = ln (21)`

We subtract ln (4) on both sides:

`xln (4) = ln (21) -ln (4)`

We divide between ln (4) on both sides:

`x = \frac {ln (21)} {ln (4)} - \frac {ln (4)} {ln (4)}\\x = \frac {ln (21)} {ln (4)} - 1\\x = 1,19615871`

Rounding:

`x = 1.1962`

Answer:

x = 1.1962