Solve the following inequality in terms of natural logarithms (ln).
(e^6x)+2 is less than or equal to 3.
same old same old
e^6x + 2 <= 3
e^6x <= 1
6x <= 0
x <= 0
see
http://www.wolframalpha.com/input/?i=plot+y%3De^%286x%29%2B2+and+y%3D3+for+-1+%3C%3D+x+%3C%3D+1
1/3
To solve the inequality (e^6x) + 2 ≤ 3 in terms of natural logarithms (ln), we can use the property of logarithms that ln(e^a) = a.
First, let's isolate the exponential term by subtracting 2 from both sides of the inequality:
(e^6x) ≤ 1
Now, let's take the natural logarithm of both sides to eliminate the exponential term:
ln(e^6x) ≤ ln(1)
Using the property mentioned earlier, we have:
6x ≤ ln(1)
Since ln(1) equals 0, the inequality becomes:
6x ≤ 0
Finally, divide both sides of the inequality by 6:
x ≤ 0/6
Simplifying, we find:
x ≤ 0
Therefore, the solution to the inequality (e^6x) + 2 ≤ 3 in terms of natural logarithms is x ≤ 0.