Question

Solve the following equation for x.
logx+log(x-3)=log(3x)
A: There are no solutions
B: x= 0,6
C: x= 6
D: x= 0

Answer 1

\[\log a + \log b = \log (ab)\]

Answer 2

\[\log_{10}x+\log_{10}(x-3) = \log_{10}3x \]

we can apply the rules of logarithms.

\[\log_{10}(x*(x-3)) = \log_{10}3x \]

\[\log_{10}(x^{2}-3x) = \log_{10}(3x)\]

\[\ln^{\log(x^{2}-3x)} = \ln^{\log{3x}}\]

this allows us to simplify this expression into something simpler

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

Answer 3

Would it then be x=0? I see that 1 would work too, but that isn't an option

Answer 4

@ingah you can't take the log(0)

Answer 5

we were able to simplify this into an easier expression

do you know how to solve this?

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

Answer 6

I tried plugging 6 in and it worked. (Earlier i mistakenly plugged in 36 instead of 6)

Answer 7

nice but let me show you how you can get to that

Answer 8

Please take a look at this @ingah

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

\[x^{2}-6x = 0\]

\[x(x-6) = 0 \]

\[\frac{ x }{ x }(x-6) = \frac{ 0 }{ x }\]

\[x-6 = 0 \]

\[x =6 \]

Answer 9

Ohhhhhhh okay. Thank you!

Answer 10

yeah so used the properties of logarithms. in the previous example.