solve for x
7^x/2=5^1-x
what i have so far is:
(x/2)ln7=(1-x)ln5
im not sure where to go from there
[ln(7) / 2] x = ln(5) - [ln(5)] x
x = ln(5) / {[ln(7) / 2] + ln(5)}
or, after some manipulation,
1/(1+log5√7)
is that simpler? who can say?
To solve for x in the equation (7^x/2) = (5^1-x), you correctly took the natural log (ln) of both sides of the equation to eliminate the exponential terms. However, there seems to be a small mistake in how you distributed the logarithm. Let me guide you through the correct steps:
1. Start with the equation: (7^x/2) = (5^1-x).
2. Take the natural logarithm of both sides of the equation: ln(7^x/2) = ln(5^1-x).
3. Apply the properties of logarithms to simplify the equation:
a. Use the power rule of logarithms: ln(a^b) = b*ln(a).
The left side becomes: x/2 * ln(7).
b. Use the power rule of logarithms again for the right side: ln(5^1-x) = (1-x) * ln(5).
Thus, the equation becomes: (x/2) * ln(7) = (1-x) * ln(5).
4. Now, let's solve for x. Distribute the x/2 on the left side:
(x/2) * ln(7) = ln(5) - x*ln(5).
5. Multiply through by 2 to clear the fraction:
x * ln(7) = 2*ln(5) - 2x*ln(5).
6. Distribute the -2x on the right side:
x * ln(7) = 2 * ln(5) - 2x * ln(5).
Now, let's isolate the x terms:
7. Move all terms involving x to one side of the equation:
x * ln(7) + 2x * ln(5) = 2 * ln(5).
8. Factor out x:
x * (ln(7) + 2 * ln(5)) = 2 * ln(5).
9. Divide both sides by (ln(7) + 2 * ln(5)) to solve for x:
x = (2 * ln(5)) / (ln(7) + 2 * ln(5)).
And that's the solution for x.