Log 3x+log 4x = log 192

First, remember that:

log

a

(

b

⋅

c

)

=

log

a

(

b

)

+

log

a

(

c

)

Therefore, we can say that:

log

(

4

x

)

+

log

(

3

x

+

1

)

=

log

(

4

x

⋅

(

3

x

+

1

)

)

⇒

log

(

12

x

2

+

4

x

)

We now rewrite the equation:

log

(

12

x

2

+

4

x

)

=

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

(

12

x

2

+

4

x

)

=

3

=>

log

10

(

12

x

2

+

4

x

)

=

3

=>

(

12

x

2

+

4

x

)

=

10

3

=>

12

x

2

+

4

x

=

1000

=>

12

x

2

+

4

x

−

1000

=

0

Let's use the quadratic formula

−

b

±

√

b

2

−

4

(

a

)

(

c

)

2

(

a

)

We now know that:

x

=

−

4

±

√

4

2

−

4

(

12

)

(

−

1000

)

2

(

12

)

We can now solve for

x

.

=>

x

=

−

4

±

√

48016

24

=>

x

=

−

4

±

4

√

3001

24

=>

x

=

−

1

±

√

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

b

step-by-step explanation:

the word superfluous means unnecessary