solve the equation
log2(x+4)-log4x=2
the 2 and 4 are lower than the g
This is what I got:
log2(x+4)+log2(4^x)=2
log2((x+4)*4^x)=2
4^x(x+4)=4
x=0 is a solution???
log2(x+4) + log4(x) = 2
log2(x+4) + log2(x)/log2(4) = 2
log2(x+4) + log2(x)/2 = 2
2 log2(x+4) + log2(x) = 4
log2((x+4)^2) + log2(x) = 4
log2(x(x+4)^2) = 4
x (x+4)^2 = 2^4
x^3 + 8x^2 + 16x − 16 = 0
x = 0.718608
To solve the given equation log2(x+4) - log4x = 2, we can start by using logarithmic properties and converting the equation into a single logarithm.
Using the property log(a) - log(b) = log(a/b), we can rewrite the equation as follows:
log2(x+4) - log4x = 2
log2(x+4) - log2(4x)/log2(4) = 2
log2(x+4) - log2(x) = 2
Next, using the property log(a) - log(b) = log(a/b), we can combine the logarithms:
log2((x+4)/x) = 2
Now, let's rewrite the equation in exponential form using the definition of logarithm:
2^2 = (x+4)/x
4 = (x+4)/x
To get rid of the fraction, we can cross multiply:
4x = x + 4
Simplifying further:
4x - x = 4
3x = 4
Divide both sides by 3:
x = 4/3
So the solution to the equation log2(x+4) - log4x = 2 is x = 4/3.