Solve for x...part deux?

Question

log_{5} x + log_{5} (x+4) = log_{5} 3

They are all Log base 5

Yes. Solve for x.

Answer 1

x(x+4)=3
x^2+4x-3=0
x=[-4+/-sqrt(16+4(1)3)]/2/1
=[-4+/-sqrt(28)]/2
=[-4+/-2sqrt(7)]/2
=0.646
(Negative is not a sol'n)

Answer 2

Use the properties log a + log b = log (ab) (in any base)

You get log_{5} x(x+4) = log_{5} 3, which means x(x+4) = 3.

Now you have a quadratic equation, x^2 + 4x - 3 = 0, that you can solve for x.

You'll get two solutions, but you must discard any negative values of x, because you have log_{5} x in your original equation, and you cannot take the log of a negative number.

Answer 3

Return to the days of slide rules where multiplying two numbers was performed by adding lengths of marking on two sticks. The problem is multiplying two terms x and x+ 4 which equals a third constant value. A quadratic Equation or Xsq + 4x-3 = 0 which has two solutions ( x-3) and (x+1) therefore x = 3 and x= -1

Answer 4

first of all the base of the Log doesnot matter sonce they all have the same base.

So your question is ? solve for x

log(x) + log(x+4) = log(3)

If my assumtion is correct let me know and i will solve this nasty beast for you.

Answer 5

logA*B=logA+logB
logA/B=logA-logB
logx+log[x+4]=log5 all base 5
x[x+4]=5
x^2+4x-5=0
x^2-5x+x-5=0
x[x-5]+1[x-5]=0
[x+1][x-5]=0
x=-1 or 5
verify
log1+log5=log5
0+log5=log5
ok

Answer 6

logx(x+4)=log3
x^2+4x-3=0
x=[-4+/-rt(16+12]/2
=[-4+/-2rt7]/2
=-2+/-rt7