Question

How do you find x in the equation log(4x)+log(3x+1)=3?

Answer 1

`x~~8.964`

Explanation:

First, remember that:
`log_a(b*c)=log_a(b)+log_a(c)`

Therefore, we can say that:
`log(4x)+log(3x+1)=log(4x*(3x+1))=>log(12x^2+4x)`

We now rewrite the equation:
`log(12x^2+4x)=3`

Now remember that:
`log(b)=log_10(b)`
And that if`log_a(b)=c`, then `b=a^c`

Therefore, we can now simplify the equation.
`log(12x^2+4x)=3`
=>`log_10(12x^2+4x)=3`
=>`(12x^2+4x)=10^3`
=>`12x^2+4x=1000`
=>`12x^2+4x-1000=0`

Let's use the quadratic formula `(-b+-sqrt (b^2-4(a)(c)))/(2(a))`

We now know that:
`x= (-4+-sqrt (4^2-4(12)(-1000)))/(2(12))` We can now solve for `x`.
=>`x= (-4+-sqrt (48016))/24`
=>`x= (-4+-4sqrt (3001))/24`
=>`x= (-1+-sqrt (3001))/6`
=>`x~~8.964,-9.197`
Now, we see that when `x<0`, whatever solution we have will be extraneous.

Therefore, `x~~8.964`