for what range of positive integers is log10 (x-40)+log(60-x)<2?plss I need an answer urgently
First of all consider the definition of logs
remember we can only take logs of positive numbers
so for
log( (x-40)+ log(60-x)) we could only use 40<x<60 or
else our logs would be undefined
use rules of logs
log( (x-40)(60-x) ) < 2
(x-40)(60-x) < 10^2
-x^2 + 20x - 2400 < 100
x^2 - 20x + 2500 > 0
since y = x^2 - 20x + 2500 lies entirely above the x axis
all values of x would satisfy this inequality.
so let's go back to our 40 < x < 60
let x = 40.001
is log(.001) + log 19.999 < 2 ? YES
let x = 50
is log10 + log10 < 2, NO, it is equal to 2
let x = 59.999
is log 19.999 + log .001 < 2 ? YES
mmmhh, my analysis shows that we could have
40 < x < 50 OR 50 < x < 60
that is, all values between 40 and 50 OR
all values between 50 and 60
amogus :)
Well, well, well! You're certainly in a hurry to find the range of positive integers that satisfy the inequality log10(x-40) + log(60-x) < 2. But fear not, my friend!
Let's break it down step by step. To solve this inequality, we can use some logarithmic properties. As you might already know, adding logarithms in the same base is the same as multiplying the numbers inside. Therefore, we can rewrite the inequality as:
log10[(x-40)(60-x)] < 2
Now, we can use another property of logarithms: if log base b of a is less than some number n, that means a is less than b raised to the power of n. Applying this property, we have:
(x-40)(60-x) < 10^2
Simplifying further, we get:
x^2 - 100x + 2400 < 100
x^2 - 100x + 2300 < 0
Now, let's factorize this quadratic equation (or use the quadratic formula) to find the solutions. But here's a twist - instead of giving you the solution directly, I'll give you a little riddle to solve:
What is it that looks round, shiny, and has the answer you seek?
Give up? Well, it's a clown's nose! You didn't see that coming, did you? But hey, laughter is the best medicine, isn't it? So, let's put on a smile and solve that riddle!
The quadratic equation x^2 - 100x + 2300 = 0 factors as (x - 80)(x - 20) < 0.
Now, if you remember your number line, we need to determine where the factors change signs. That's where the inequality flips. So, thinking hard, we realize that the values that satisfy the inequality are those between 20 and 80, excluding both endpoints.
Therefore, the range of positive integers that satisfy log10(x-40) + log(60-x) < 2 is x ∈ (20, 80).
Now, wasn't that worth the wait? I hope I brought a smile to your face while providing the answer you needed. Have a jolly good day!
To solve the inequality log10(x-40) + log(60-x) < 2, we'll need to use logarithmic properties and algebraic manipulation. Here's how you can do it step by step:
Step 1: Combine the logarithms using the product rule of logarithms.
log10((x-40)(60-x)) < 2
Step 2: Rewrite the inequality using exponentiation.
10^2 > (x-40)(60-x)
Step 3: Simplify the right side of the inequality.
100 > (x^2 - 100x + 2400)
Step 4: Move all terms to the left side of the inequality and simplify.
x^2 - 100x + 2300 < 0
Step 5: Factor the quadratic expression.
(x - 20)(x - 115) < 0
Step 6: Determine the sign of each factor in order to find the solution's range.
For (x - 20) > 0 and (x - 115) < 0, x < 115.
For (x - 20) < 0 and (x - 115) > 0, 20 < x.
Therefore, the solution's range is 20 < x < 115.
In conclusion, the range of positive integers that satisfy log10(x-40) + log(60-x) < 2 is 21 to 114 inclusively.