SOLUTION: We were asked to solve logarithmic inequations, and although I could solve it, I stopped at a dead end, in which I couldn't find a ''good'' answer..
i) 2*ln(x+2)-1 > 0
ii) l
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Question 1028170: We were asked to solve logarithmic inequations, and although I could solve it, I stopped at a dead end, in which I couldn't find a ''good'' answer..
i) 2*ln(x+2)-1 > 0
ii) log(3-2x) > 1 + log(x+5)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i would solve for the equality and then determine the inequality.
the first equation is 2 * ln(x+2) - 1 > 0
set that equal to 0 to get 2 * ln(x+2) - 1 = 0
add 1 to both sides of the equation to get 2 * ln(x+2) = 1
divide both sides of the equation by 2 to get ln(x+2) = 1/2
this is true if and only if e^(1/2) = x + 2
solve for x to get x = e^(1/2) - 2 which results in x = .3512787293
when x is greater than this, you will find that the equation is positive.
when x is less than this, you will find that the equation is negative.
since it's only equal to 0 at one point, this means that all values of x less than -.3512787293 will give you a negative result and all values of x greater than -.3512787293 will give you a positive result.
therefore, the solution is that 2*ln(x+2)-1 > 0 when x > .3512787293
here's the graph of the equation.
your second equation is log(3-2x) > 1 + log(x+5)
subtract log(x+5) from both sides of the equation to get:
log(3-2x) - log(x+5) > 1
since log(a) - log(b) is equal to log(a/b), your equation becomes:
log((3-2x)/(x+5)) > 1
set the equation equal to 1 to get:
log((3-2x)/(x+5)) = 1
this is true if and only if 10^1 = (3-2x) / (x+5)
simplify to get 10 = (3-2x) / (x+5)
multiply both sides of this equation by (x+5) to get:
10 * (x+5) = 3-2x
simplify to get 10x + 50 = 3 - 2x
add 2x to both sides of this equation and subtract 50 from both sides of this equation to get:
12x = -47
divide both sides of this equation by 12 to get:
x = -47/12
if you replace x with -47/12 in the original equation, you will see that the equation is true.
your original equation is log(3-2x) > 1 + log(x+5)
this equation will be valid as long as 3-2x is > -0 and as long as (x+5) is > 0
otherwise the equation becomes invalid because you can't take the log of a number that is not positive.
for 3 - 2x to be positive, then solve 3 - 2x > 0
add 2x to both sides of the equation to get 3 > 2x
divide both sides of this equation by 2 to get 3/2 > x
this means that x has to be < 3/2 or 1.5.
if x is greater than or equal to 1.5, the equation is invalid.
likewise x + 5 must be > 0
start with x + 5 > 0
subtract 5 from both sides of this equation to get x > -5
you have x must be greater than -5 and less than 1.5
you know that the equation is equal to 0 when x = -47/12
when x = 0, the equation of log(3-2x) > 1 + log(x+5) becomes:
log(3) > 1 + log(5)
this results in log(3) smaller than 1 + log(5) so the equation is negative when x > -42/12.
when x = -4, you get log(3-2x) > 1 + log(x+5) becomes:
log(11) > 1 + log(1) which is true.
not that x can't be less than or equal to -5 and can't be greater than or equal to 3/2.
the solution is therefore that:
log(3-2x) > 1 + log(x+5) when x > -5 and x < 47/12.
the solution is therefore -5 < x < 47/12
if you start with log(3-2x) > 1 + log(x+5) and subtract the expression on the right side of the equation from both sides of the equation you will get:
log(3-2x) - log(x+5) - 1 > 0
graph the equation of y = log(3-2x) - log(x+5) - 1 and look at the graph.
the graph is shown below:
best i can do for now because i ran out of time.
if you have further questions, send me an email.
hopefully this will help.
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