To solve the equation ln(x) + 6ln(2x) = 1, we will start by simplifying the equation using properties of logarithms.
Recall that ln(a) + ln(b) = ln(ab). We can apply this to the second term:
ln(x) + 6ln(2x) = ln(x) + ln((2x)^6)
Now, we can rewrite (2x)^6:
(2x)^6 = 2^6 * x^6 = 64x^6. Therefore, we can combine the logarithms:
ln(x) + ln(64x^6) = ln(64x^7)
So now, our equation reads:
ln(64x^7) = 1
Next, we will exponentiate both sides to eliminate the logarithm:
e^(ln(64x^7)) = e^1
This simplifies to:
64x^7 = e
Now we’ll solve for x:
x^7 = e / 64
Thus, we take the seventh root:
x = (e / 64)^(1/7)
For a numerical approximation, we can substitute the value of e (approximately 2.71828):
x ≈ (2.71828 / 64)^(1/7)
x ≈ (0.0425)^(1/7)
Using a calculator, this gives:
x ≈ 0.7513
In conclusion, the value of x in the equation ln(x) + 6ln(2x) = 1 is approximately 0.7513.