Question

Solve the equation below for x. log(5x) + log(2x) = 1 O A. x = 10/7 O B. x = 1; x = -1 O c. x= 1 O D. There is no solution.

Answer 1

So we need to solve the following equation:

`\log (5x)+\log (2x)=1`

There are a few properties of logarithmic functions that we should remember. First, the logarithm of a negative number doesn't exist which means that x must be a positive number. Second, the addition of logarithms meets the following property:

`\log (a)+\log (b)=\log (a\cdot b)`

If we apply this to our equation we get:

`\log (5x)+\log (2x)=\log (5x\cdot2x)=\log (10x^2)=1`

Now we can pass the logarithm to the right side of the equation:

`\begin{gathered} \log (10x^2)=1 \\ 10x^2=10^1=10 \\ 10x^2=10 \\ x^2=1 \end{gathered}`

There are two possible solutions for x^2=1. These are x=1 and x=-1, however as I stated before x can't be a negative number which means that the solution of the equation is:

`x=1`

Then option C is the correct one.