You can put this solution on YOUR website! log(x+3)+log(x) = 1
Solving equations like this usually start with transforming it into one of the following general forms:
log(expression) = number
or
log(expression1) = log(expression2)
Since your equation has the non-log term of 1, transforming it into the all-log second form will be difficult. So we will aim for the first form. To achieve the first form we only have to find a way to combine the two logs into one.
The two logs are not like terms so we cannot just add them. (Like logarithmic terms have the same bases and the same arguments. Your logs have the same base, 10, but the arguments are different.)
Fortunately there is a property of logarithms:
which allows us to combine two logs if the bases are the same and the coefficients are 1's. Your logs have the same bases and their coeficients are 1's so we can use this property to change
log(x+3)+log(x) = 1
into
log((x+3)*(x)) = 1
which simplifies to:
We now have the first form.
The next step with the first form is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:
which simplifies to:
The logarithms are now gone. And we have a quadratic equation in its place. Next we solve this equation. (We will use factoring.)
0 = (x+5)(x-2)
x+5 = 0 or x-2 = 0
x = -5 or x = 2
Last of all we check. This is not optional! Solutions to logarithmic equation must be checked to ensure that all bases and arguments of the original equation are valid. (Valid bases are positive but not 1 and valid arguments are positive.
Checking x = -5:
log((-5)+3)+log(-5) = 1
As we can see, both arguments will be negative. These are invalid so we reject this solution. (Note: If only one argument was negative we would still reject the solution.)
Checking x = 2:
log((2)+3)+log(2) = 1
With this solution, both arguments are valid. So this solution checks out OK.
So there is only one solution to your equation: x = 2.
Note: Please post your questions in a relevant category. This was neither a word problem nor a linear equation/system.
You can put this solution on YOUR website! The log.s are to base 10 as I understood, & if so,
first, 1=log10; So, we've:
log(x+3) + log(x) = log10;
=> using the property of logarithm,
loga+logb=log(a*b),
log[(x+3)(x)]=log10;
Now we can simply simplify and solve this: x^2+3x-10=0;
factorize
=> (x-2)(x+5)=0
=> x=-5 or x=2. But x=-5 isn't in the domain & can't be answer.
You can simply check by using the properties or like this: log(2+3)=log5=0.6990 and log2=0.3010
=> log5 + log 2= 1.
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