Question

What is the solution to 4^(log4(x+8))=4^2x=-8x=-4x=4x=8

Answer 1

Answer:

x=8

Step-by-step explanation:

Use the main property of logarithms:

$a^{\log_ab}=b.$

Then $4^{\log_4(x+8)}=x+8.$

Now the equation takes look

$x+8=4^2,\\ \\x+8=16,\\ \\x=16-8,\\ \\x=8.$

Check whether x=8 is the solution:

$4^{\log_4(8+8)}=4^{\log_416}=4^{\log_44^2}=4^{2\log_44}=4^{2\cdot 1}=4^2=16.$

Answer 2

Hi there! The answer is x = 8.

${4}^{ log_{4}(x + 8) } = {4}^{2}$
Let's solve this equation step by step!


First we use the following rule.
${g}^{a} = {g}^{b} \\ a = b$
which basically means that an exponential equation with the same base on both sides of the equation must also have the same exponents.

We get the following.
$log_{4}(x + 8) = 2$

Now we use the following rule for logarithms.
$log_{a}(x) = b \\ x = {a}^{b}$

When we apply this rule in our eqiation, we end up with the following.
$x + 8 = {4}^{2}$

Now we only need to do some basic math, work out the exponents first.
$x + 8 = 16$

Subtract 8.
$x = 8$

~ Hope this helps you!