Solve log[base 2] x+log[base 2](X+6)=log[base 2]16.?

Question

Please Help! :)

Answer 1

log[base 2] x+log[base 2](x+6)=log[base 2]16


2^{log[base 2](x) + log[base 2](x+6)} = 2^{log[base 2](16)}

2^log[base 2](x) * 2^log[base 2](x+6)} = 2^{log[base 2](16)}

x*(x+6) = 16

x² + 6x - 16 = 0

(x+8)(x-2) = 0

x=-8 x=2

Since the log of x<0 makes no sense

x=2 is the only solution

Answer 2

Log (2) x(x+6) = log(2) 16

If all of these are log to the base 2, make everything 2^(some power).

2^log(2) x(x+6) = 2^log(2) 16

x^2 + 6 x = 16

x^2 + 6 x = 16

x^2 + 6 x - 16 = 0

Factor this:

(x - 2) (x + 8) = 0

x = 2 or -8

Since all these logarithms have the same base, the logarithm part can almost be entirely ignored by taking 2 to the power of the expression.

Answer 3

use log rule that announces loga +logb=log(ab) iff the logs have a similar base. then you definately get log base 4 (x(x-6))=2 components of logs if log base a of b=c then a^c=b on your case 4^2=x(x-6) Simplifying we get x^2-6x-sixteen=0 (x-8)(x+2)=0 so x=8 or x=-2 no longer finished yet! Plug the two in to determine in the event that they paintings. x=-2 won't paintings in view that log base 4 of -2 is UNDEFINED as a results of actuality THE IMPUTS OF A LOG could be STRICTLY greater beneficial than 0 relish

Answer 4

All logs are to base 2.
log x+log (x+6) = log 16
log(x(x+6) = log 16
x^2+6x = 16
x^+6x-16=0
(x+8)(x-2) = 0
x =2
x=-8 is rejected

Answer 5

this is a cool problem...
well, they are all cool to me..;).. here is the deal..

recall how to contract the logs using the log rules.. the sum of the logs = log of the products..

to get..
log[bse]2 of x(x+6) = 4..

since log bse2 of 16 is 4

then.. x(x+6)= 2 to the 4... remembering base is 2... exponent four..

you get a quadratic equation..
x^2 +6x =16

here, you can either factor or complete the square,,
I'll complete the square for fun..;)

you get x^2 +6x +9 = 16 +9..

factoring the left side to get
(x+3)^2 = 25.. as you can see.. both sides are perfect squares..

x=-3 +- 5
to obtain.. 2 and, -8..

good luck..but since -8 gives a negative log,,and does not exist. we keep the plus 2.

Answer 6

Okay, so first of all we should remember that having two logarithms of the same base added means that we can multiply what's inside the logarithms together in one single logarithm. That is:

log[2] (x) + log[2] (x+6) = log[2] (16)
log[2] (x(x+6))=log[2] (16)

If we take away the logs:
x(x+6)=16
x^2 + 6x -16=0
(x+8)(x-2)=0
x=-8, x=2

Answer 7

log[base 2](x) + log[base 2](x + 6) = log[base 2](16)

First, combine the logs on the left hand side using the identity log(a) + log(b) = log(ab).

log[base 2](x(x + 6)) = log[base 2](16)

Take the antilog of both sides; this effectively equates the logs' arguments.

x(x + 6) = 16

Solve as normal.

x^2 + 6x = 16
x^2 + 6x - 16 = 0
(x + 8)(x - 2) = 0

Therefore,
x = {-8, 2}
However, for logarithmic equations, we must check for extraneous solutions. Plug each value into the original equation; note that we cannot take the log of a negative number.

If x = -8, we can see that this is immediately invalid because log[base 2](-8) is invalid. Reject x = -8.

If x = 2, we don't take the log of a negative number, so this one we keep.

Therefore,
x = 2 is the only solution.

Answer 8

log[base 2] (x(x+6))=log[base 2] (16)
x(x+6)=16
x^2+6x-16=0
x^2+8x-2x-16=0
x(x+8)-2(x+8)=0
(x+8)(x-2)=0
x=-8,2
as log cannot be -ive therefore x has only one value i.e.
x=2

Answer 9

x^2+6x - 16 = 0, x>0
x = 2