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

x=-8
x=-4
x=4
x=8

User Colin Fine
by
8.2k points

2 Answers

4 votes
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!


User Bitluck
by
7.2k points
0 votes

Answer:

x=8

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.

User LordNeo
by
6.8k points