Solve to the nearest thousandth.
5^1+x = 2^1-x
To solve for x in the equation 5^(1+x) = 2^(1-x), we will first take the logarithm of both sides of the equation.
Let's start by taking the natural logarithm (ln) of both sides:
ln(5^(1+x)) = ln(2^(1-x))
Now, according to the logarithm rule, we can bring down the exponent in front of the logarithm:
(1+x) ln(5) = (1-x) ln(2)
Next, we can expand the equation further by distributing the logarithms:
ln(5) + x ln(5) = ln(2) - x ln(2)
Now, let's collect all the x terms on one side and the constant terms on the other side:
x ln(5) + x ln(2) = ln(2) - ln(5)
Using the logarithm properties, we can simplify this equation:
x(ln(5) + ln(2)) = ln(2) - ln(5)
Now we can divide both sides of the equation by (ln(5) + ln(2)):
x = (ln(2) - ln(5)) / (ln(5) + ln(2))
Now, we can approximate this value by using a calculator:
x ≈ -0.191
Therefore, to the nearest thousandth, x is approximately -0.191.
(1+x)*Log5 = (1-x)*Log2.
Divide both sides by Log5:
1+x = (1-x) * 0.431.
1+x = 0.431-0.431x.
1.431x = -0.569.
X = -0.398.