Question

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

Answer 1

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

Answer 2

Answer: $x$≈$1.196$

Step-by-step 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:

 $log_b(x) = \frac{log_a( x)}{log_a(b)}$

Then you get:

$x =\frac{log(21)}{log(4)}-1$

$x$≈$1.196$