solve the equation:
log base 10 of x - log base 10 of (7-4x)=2
Please help i have been stuck on this for awhile!
x/(7-4x) = 10^2
x = 100(7-4x)
x = 7400 -400x
401x = 7400
x = 7400/401
But in log (7-4x)
since we can only take logs of positives,
7-4x > 0
-4x > -7
x< 7/4 or x < 1.75
since 7400/401 > 1.75
there is no solution to this equation.
actually, the typos above conceal the fact that
x = 100(7-4x)
x = 700 - 400x
x = 700/401 which is in fact less than 7/4
Egg on face
(intentions were honourable)
To solve the equation log base 10 of x - log base 10 of (7-4x) = 2, we can use the logarithmic properties to simplify the equation.
Logarithmic Property 1: log base a of b - log base a of c = log base a of (b/c)
Using this property, we can rewrite the equation as:
log base 10 of (x/(7-4x)) = 2
Next, we need to convert the logarithmic equation into an exponential equation. The exponential form of a logarithmic equation is given by:
a^b = c if and only if log base a of c = b
Using this equation, we can rewrite the equation as:
10^2 = x/(7-4x)
Simplifying further:
100 = x/(7-4x)
Now, we have a simple algebraic equation to solve. Let's solve it step by step:
1. Multiply both sides of the equation by (7-4x) to eliminate the denominator:
100(7-4x) = x(7-4x)
Simplifying both sides gives:
700 - 400x = 7x - 4x^2
2. Rearrange the equation to bring all terms to one side:
4x^2 - 7x + 700 = 0
3. This equation can be solved using factoring, completing the square, or the quadratic formula. In this case, we will use the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation (ax^2 + bx + c = 0):
x = (-(-7) ± √((-7)^2 - 4(4)(700))) / (2(4))
Simplifying further:
x = (7 ± √(49 + 11200)) / 8
x = (7 ± √11249) / 8
4. We can simplify the square root:
√11249 = √(11 * 11 * 101) = 11√101
Therefore, we have two solutions for x:
x1 = (7 + 11√101) / 8
x2 = (7 - 11√101) / 8
These are the solutions to the given equation.