Solve for x:
log (x^4)=(log x)^2
Note, there are 2 solutions, A and B, where A<B.
A=
B=
let z = log x
4 z = z^2
z = 4 or z = 0
log x = 4 or log x = 0
10^log x = x = 10^4 = 10,000
or
10^log x = x = 10^0 = 1
To solve the equation log(x^4) = (log x)^2, we can rewrite it using the properties of logarithms.
First, recall that the power rule of logarithms states that log(a^b) = b * log(a).
Using this rule, we can rewrite the equation as 4 * log(x) = (log x)^2.
Let's introduce a substitution to make solving for x more manageable. Let's define a new variable y as log(x).
Substituting y into the equation, we have 4y = y^2.
Now, we have a quadratic equation. We can rearrange it and set it equal to zero:
y^2 - 4y = 0
Factoring the equation, we get:
y(y - 4) = 0.
From the factored form, we can see that there are two possible solutions: y = 0 and y - 4 = 0.
Solving for y in each equation separately:
1. y = 0:
Since y = log(x), if y = 0, then log(x) = 0.
Using the definition of logarithms, 10^0 = x, so x = 1.
2. y - 4 = 0:
Solving for y, we get y = 4.
Using the definition of logarithms, 10^4 = x, so x = 10,000.
Therefore, the two solutions to the equation are x = 1 and x = 10,000.
Thus, A = 1 and B = 10,000.