Question

Which of the following is a solution to the equation log 4 x + log 4 (x – 3) = 1 ?

x = 6/5

x = 2

x = –3

x = 4

Answer 1

What you want to do first is basically simplify the equation: 4 x + log 4 (x – 3) = 1. Can you do that?

Answer 2

hold on let me work it out

Answer 3

okay :)

Answer 4

Because the logs are of the same base, you can combine them together. An addition of logs of the same base allows to multiply the contents of those logarithms inside of one log. So this means we can rewrite the addition of your logs as:

\[\log _{4}[x(x-3)]=1\] Now that we have this, we use a second logarithm property that says:

\[\log _{b}M = N\]is the same as:
\[b ^{N}=M\]

So in this case, b is 4, N is 1 and M is x(x-3) So that being said, we can write this:

\[4^{1}=x(x-3)\]From here, we need to move everything to one side of the equation, meaning we subtract both sides by 4 to get:

\[x(x-3) -4 = 0\]If I multiply this out, I have a quadratic equation:

\[x ^{2}-3x - 4=0\]

From here we want to factor this quadratic and solve for x. In this case, we can factor the quadratic into (x-4)(x+1)

When you have the factors of a quadratic, you set each factor equal to 0 and then solve for x. So doing that means we have:

x-4 = 0
x + 1 = 0 therefore
x = 4 and x = -1

So those are two answers we get. Now just because we found these values of x doesnt mean theyre actual answers, we have to check them. So we want to replace x in our original equation with 4 and then replace x in our original equation with -1. So starting with 4, we get:

\[\log _{4}(4)+\log _{4}1= 1\]And if I simplify that, logbase 4 of 4 is 1 and log base 4 of 1 is 0. Therefore we get 1 = 1 which is true. Now we would plug in -1:

\[\log _{4}(-1) + \log _{4}(-4) = 1\]Now we are not allowed to have a negative inside of a logarithm, so that means only x = 4 is an answer. And feel free to ask questions about any of what I did : 3

Answer 5

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Answer 6

well that made it easy for me

Answer 7

thanks a lot

Answer 8

Yeah, sure.