Solve the equation log(x^2+4)=2+logx-log20
AAAaannndd the bot gets it wrong yet again!
log(x^2+4)=2+logx-log20
log(x^2+4)-logx+log20 = 2
20(x^2+4)/x = 100
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 or 4
check x=1:
log5=2+log1-log20
log5+log20 = 2
log100 = 2
yes
check x=4:
log20=2+log4-log20
2log20 = log400
yes
Apologies for the incorrect solution. You are correct.
The correct solutions to the equation log(x^2+4)=2+logx-log20 are x = 1 and x = 4. Thank you for pointing out the mistake.
Thanks oobleck, saved me some puzzlement.
You're welcome! I'm glad I could help. If you have any more questions or need further assistance, feel free to ask.
ROTFL
I'm glad I could bring some laughter to your day! If you have any more questions or need assistance with anything else, feel free to ask.
To solve the equation log(x^2+4) = 2 + log(x) - log(20), we can simplify the equation first.
Using the properties of logarithms, we can rewrite the equation as:
log[x^2+4 / (20 * x)] = 2
Now, let's exponentiate both sides of the equation using the base 10:
10^(log[x^2+4 / (20 * x)]) = 10^2
[x^2+4 / (20 * x)] = 100
Multiply both sides of the equation by (20 * x):
x^2 + 4 = 100 * (20 * x)
x^2 + 4 = 2000x
Rearrange the equation to one side:
x^2 - 2000x + 4 = 0
This is a quadratic equation. To solve it, we can either factor it or use the quadratic formula. In this case, factoring is not easy or apparent, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 1, b = -2000, and c = 4. Substituting these values into the formula, we get:
x = (-(-2000) ± √((-2000)^2 - 4 * 1 * 4)) / (2 * 1)
Simplifying further:
x = (2000 ± √(4000000 - 16)) / 2
x = (2000 ± √(3999984)) / 2
x = (2000 ± 1999.996) / 2
Solving for x using the plus and minus roots:
x1 = (2000 + 1999.996) / 2 ≈ 1999.998
x2 = (2000 - 1999.996) / 2 ≈ 0.002
Therefore, the solutions to the equation log(x^2+4) = 2 + log(x) - log(20) are approximately x1 ≈ 1999.998 and x2 ≈ 0.002.