solve log(x+5) + log(x+2) = 1?

The above can also be written as log[(x+5 )*(x+2)]=1
next just make each side the exponent of 10 which will simplify to (x+5)(x+2)=10....x^2+7x+10=10....x^2+7x=0...x(x+7)=0....x=0 or x=-7. The x=-7 solution does not work, however, because it is not in the domain: for example, log(x+5) would become log(-2) which is not defined. Therefore, the only solution is x=0.

2007-01-25 06:34:04 · answer #1 · answered by bruinfan 7 · 3⤊ 0⤋

The answer is 0 because
log 5 + log 2 = log 10 = 1

2007-01-25 06:36:58 · answer #2 · answered by ironduke8159 7 · 0⤊ 1⤋

log(x+5)+ log(x+2)=1

log((x+5)(x+2)) =1 If we are working with log in base 10 we get

(x+5)(x+2)=10====> x^2 +7x=0 so x= 0 or x= -7 But the last can´t be because x+5 and x+2 would be negatve and log of
negative numbers does NOT exist
So your only solution is x=0

2007-01-25 07:30:10 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋

The "trick" is using rules of exponents and logs.

1) log(x) + log(y) = log(xy) so ...

log(x+5) + log(x+2) = log[(x+5)(x+2)] = log[x^2 + 7x +10]

note: I use ^ to mean raise to the power of so x^2 is x "squared."

2) Exponentiate (make both sides exponents of 10)

10^log(x) = x so ....

10^log[(x+5)(x+2)] = 10^1

Simplifies to ...

(x+5)(x+2) = 10 Now you have a quadratic equation to solve

In the end I get x = 0 or -7

2007-01-25 07:14:58 · answer #4 · answered by profsteves 1 · 0⤊ 0⤋

we know that:
logab=loga+logyb
for a=x+5 and b=x+2
then, for log(x+5)+log(x+2)=1
set; X^2-SX+P=1 (1)
or S=a+b and P=a*b
then S=(x+5) +(x+2) or S=2x+7 and P=(x+5)(x+2) or P=x^2+7x+10
equation (1) becomes:
X^2-(2X+7)X+X^2+7X+10=1
X^2-2X^2-7X+X^2+7X+10=1
0x=1-10 0x=-9 impossible
no value for x
S={ } or S is empty

2007-01-25 07:13:31 · answer #5 · answered by Johnny 2 · 0⤊ 0⤋

log generally implies base 10 logarithm.

If x = 0 then log(5) + log(2) = 1

if x = -7 then log(-2) + log(-5) = 1 How do you define log of negative numbers?

2007-01-25 06:49:02 · answer #6 · answered by John W 3 · 0⤊ 1⤋

log(x+5) + log(x+2) = 1
log(x+5)(x+2)=1 (property of log)
log_10 (x+5)(x+2)=1 (log has base of then when the base is not indicated)

(x+5)(x+2)=10^1
x^2+7x+10=10
x^2+7x=0
x(x+7)=0
x= 0 or -7

2007-01-25 06:30:02 · answer #7 · answered by 7 · 1⤊ 1⤋

log(x+5) + log(x+2) = 1
log((x+5)(x+2)=1
(x+5)(x+2)=10
x^2+7x+10=10
x^2+7x=0
x=0
x+7=0
x=-7
x=0, -7

2007-01-25 06:30:02 · answer #8 · answered by yupchagee 7 · 0⤊ 1⤋

a million/2*log(4x+5)=log x; log(4x+5)^(a million/2)=log x; log ((4x+5)^(a million/2))/x)=0; ((4x+5)^(a million/2))/x=a million; 4x+5=x^2; x^2-4x-5=0; via fixing the above equation you get your answer...............

2016-12-16 13:24:57 · answer #9 · answered by ? 4 · 0⤊ 0⤋

solve log(x+5) + log(x+2) = 1?

9 answers