Log4(x^2+3x)-log4(x+5)=1?

Question

the log 4 means log base 4

i didnt know how to make it little.

thanks so much

Answer 1

Log(a)-Log(b) = Log(a/b)

log4 (x^2+3x) - log4 (x+5) =1

log4 (x^2+3x)/ (x+5) = 1

(x^2+3x)/(x+5) = 4^1

x^2+3x=4(x+5)

x^2+3x=4x+20

x^2+3x-4x-20=0

x^2-x-20=0

x^2-5x+4x-20=0

x(x-5)+4(x-5)=0

(x-5)(x+4)=0

x=5 or x=-4

Answer 2

log4 [(x^2 + 3x)/(x+5)] = 1 (logx-logy=log(x/y))

(x^2 + 3x)/(x+5) = 4^1 (log(base a) x=n => x=a^n)

This is a quadratic equation. You should be able to solve.

Answer 3

log_4 [(x^2+3x)/(x+5)]=1

(x^2+3x)/(x+5)=4

x^2+3x-4x-20=0

x^2-x-20=0

(x-5)(x+4)=0

x=5 or x=-4

Proving:

log_4(5^2+15)-log_4(5+5)=log_4(40/10)=... 4=1 OK

For x=-4

log_4(16-12)-log_4(1)=1+0 =1 OK

Bye

Answer 4

First collect the log.

log4 [ (x^2+3x) / (x+5) ] = 1

Then use the change of base law.

{ ln [ (x^2+3x) / (x+5) ] } /  ln4 = 1

Take the ln 4 to the other side

ln [ (x^2+3x) / (x+5) ]  = ln4

Exponentials of both sides

(x^2+3x) / (x+5) = 4

Take the x+5 to the other side

x^2+3x = 4x + 20

Collect the terms and form a quadratic

x^2 - x - 20 = 0

Factorise

(x - 5) (x + 4) = 0

x = 5 or -4.

Answer 5

We know that if the base is the same:

log a + log b = log (a*b)

log a- log b= log (a/b)

So we can simplify it to log 4 ((x^2+3x)/(x+5))=1

we know that if log b a =c, b^c=a

so 4^1=(x^2+3x)/(x+5). And then, we multiply both sides with (x+5) with consideration that x+5 mustn't =0.

so we get 4(x+5)=x^2+3x

x^2-x-20=0

(x-5)(x+4)=0

so x=5 or x=-4.

because both result is not -5, we can use both, but we also have to consider the condition of the problem.

log a has result if a>0. So x^2+3x>0

x(x+3)>0

with using analysis, we will get that x must <-3 or >0 , so both result can still be used. The last criteria is that (x+5)>0, x>-5.Both results also satisfy this criteria. So we can conclude that x=5 or x=-4

Answer 6

hi there...

just write log b4 = it means base 4.

here...

use log a - log b = log (a/b)

then...

log b4 (x^2 + 3x)/(x+5) = 1

then...

4^1 = (x^2 + 3x)/(x+5), do cross multiply

4(x+5) = x^2 + 3x

4x + 20 =x^2 + 3x

x^2 + 3x - 4x - 20 = 0

x^2 - x - 20 = 0

(x-5)(x+4)=0

x = 5 or x = -4

done ..see ya..

Answer 7

log4(X^2=3X0-log4(X+5)=log4(4)

log4{ (X^2+3X) / (X+5) } = log4(4)

cut down the log 4 on both side

X^2+3X=4X+20 --> X^2-X-20=0

quadratic equation(use ur calculator if u don't know how to factor it) then

(X-5)(X+4)=0

X= 5 , X=-4

Answer 8

first you use the division property to make the two logs into a single log.

log4((x^2+3x)/(x+5))=1

Now we define 1 as a logarithm of base 4. (4^1= 1 as a logarithm of base 4)

log4((x^2+3x)/(x+5))=log4(4)

From this statement, we can conclude the following:

(x^2+3x)/(x+5)=4 (Multiply both sides of the equation by (x+5)

x^2+3x=4x+20 (set the equation equal to zero)

x^2-x-20=0 (factor)

(x-5)(x+4)=0

x=5,x= -4 (check for domain errors by substituting in the values for x)

Both check, so your answers are x=5, x=-4