SOLUTION: please help solve the following problem for the varible x:
i know the answer is 4 but i don't know how to get it.
log(x+21)+log(x)=2
Algebra.Com
Question 78144: please help solve the following problem for the varible x:
i know the answer is 4 but i don't know how to get it.
log(x+21)+log(x)=2
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given:
.
log(x+21)+log(x)=2
.
By the rules of logs, when you add two logs it is the same as taking the log of their
products. In other words:
.
log(A) + log(B) = log(A*B)
.
We can make use of this product rule to re-write the left side of the equation as:
.
log[(x+21)*(x)] = 2
.
do the multiplication inside the brackets on the left side to get:
.
log[x^2 + 21x] = 2
.
Convert this base 10 log to its exponential form to get:
.
[x^2 + 21x ] = 10^2
.
But since 10^2 = 100, the right side can be changed and the problem becomes:
.
x^2 + 21x = 100
.
subtract 100 from both sides and you get:
.
x^2 + 21x - 100 = 0
.
The left side of this equation factors to become:
.
(x + 25)*(x - 4) = 0
.
This equation will be true if either of the factors equals zero because a multiplication
by zero on the left side will make the left side zero and equal to the right side.
So either:
.
x + 25 = 0 which after subtracting 25 from both sides becomes x = -25
.
or x - 4 = 0 which after adding 4 to both sides becomes x = +4
.
But return to the original problem and note that if x = -25 the problem becomes:
.
log(-4) + log(-25) = 0
.
You cannot have the log of a negative number ... so x = -25 is an invalid answer and we
disregard it.
.
So the answer we got is x = +4.
.
As a check, substitute +4 for x in the original problem and you get:
.
log(4 + 21) + log(4) = 2
.
which simplifies to:
.
log(25) + log(4) = 2
.
use a calculator to get each of the two log terms on the left and you get:
.
1.397940009 + 0.602059991 = 2
.
That checks out ... so the answer x = 4 is correct!
.
Hope this helps you to understand a couple of the properties of logarithms and how to
use them to your advantage in working problems such as this one. In particular, we used
the fact that the addition of logs can be treated as the log of their products and also
the conversion of logs to exponential form by raising the base of the logs to the power
of the right side and setting that equal to the quantity we are taking the log of.
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