Xlog 1.0.1

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When you enter an equation into the calculator, the calculator will begin by expanding (simplifying) the problem. Then it will attempt to solve the equation by using one or more of the following: addition, subtraction, division, taking the square root of each side, factoring, and completing the square.

Apr 14, 2013 If it is log for both denominator and numerator then write log(1+ax) -log(1+x^2) use u.v rule in both and integrate u get after applying limits is.

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Algebra - Logarithm Solvers, Trainers and Word Problems - SOLUTION: log(2x)-log(x-3)=1 Log On Algebra: Logarithm Section. Solvers Solvers. Lessons Lessons.
We need to reject any solution that makes an argument to any logarithm zero or negative. Always use the original equation to check. Checking x = -2: which simplifies to: As we can see, both of the logarithms have negative arguments. So we must reject this solution. (If even one argument was zero or negative we would still reject this solution.).

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For addition and subtraction, use the standard + and - symbols respectively. For multiplication, use the * symbol. A * symbol is not necessary when multiplying a number by a variable. For instance: 2 * x can also be entered as 2x. Similarly, 2 * (x + 5) can also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5). The * is also optional when multiplying with parentheses, example: (x + 1)(x - 1).

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X Log10x Derivative

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Algebra -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: solve: logx + log(x-3) = 1 Log On

X Log 1.0.1 1

Question 272775: solve:
logx + log(x-3) = 1

Answer by jsmallt9(3757) (Show Source):

You can put this solution on YOUR website!

To solve equations where the variable is in the argument of a logarithm, like this one, you usually start by transforming the equation into one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)
Since your equation has the non-logarithmic term of 1 (on the right side of the equation), it will be more difficult to achieve the all-logarithm second form. So we will aim for the first form.
For the first form we need one side of the equation to be a single logarithm (without a coefficient). Your equation has two logarithms on the left so we need to combine them into one somehow. The logarithms are not like terms so we cannot add them. But there is a property of logarithms, , which allows us to combine the two logarithms into one:
which simplifies to:
We have now achieved the first form. With this form we proceed by rewriting the equation in exponential form:
which simplifies to:
We have eliminated the logarithms entirely! We now have a quadratic equation to solve. We start by getting one side equal to zero (by subtracting 10 from each side):
Next we factor:
From the Zero Product Property we know that this product is zero only if one the of the factors is zero:
or
Solving these we get:
or
When solving logarithmic equations it is important, not just a good idea, to check your answers. We need to reject any solution that makes an argument to any logarithm zero or negative. Always use the original equation to check.
Checking x = -2:
which simplifies to:
As we can see, both of the logarithms have negative arguments. So we must reject this solution. (If even one argument was zero or negative we would still reject this solution.)
Checking x = 5:
which simplifies to:
As we can see, both of the logarithms have positive arguments. So we have no reason to reject this solution. (You're welcome to finish the check.)
So the only solution to your equation is x = 5.

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