Mastering the Mysteries of Logarithms: Solving for x in a Tricky Equation

[DeanBH]

i've got a logashizm problem to this point

the log is base 2log (x(x+3)^2 / (4x+2)) = 1

apparently x(x+3)^2 / (4x+2) = 2

no idea why, halp?

thxgod damn it, nvm

2^1 = (x(x+3)^2 / (4x+2)) when you take the damn log out.

 

[BrendanH]

exactly

 

[Architeuthis Dux]

I can understand your frustration with this logashizm problem. Logarithms can be tricky and require a thorough understanding of their properties and rules to solve equations like this. In this case, the log is base 2, which means we are working with binary logarithms.

To solve this equation, we first need to understand that taking the log of both sides of an equation cancels out the exponent. Therefore, we can rewrite the equation as 2^1 = x(x+3)^2 / (4x+2). This simplifies to 2 = x(x+3)^2 / (4x+2).

To further simplify this equation, we can use the distributive property and expand (x+3)^2 to get 2 = (x^2 + 6x + 9) / (4x+2). We can then cross-multiply and solve for x to get x = 1 or x = -3.

I understand that this may seem confusing, but with practice and a deeper understanding of logarithms, you will be able to master these types of equations. Keep in mind the properties of logarithms, such as log(a*b) = log(a) + log(b) and log(a/b) = log(a) - log(b), and always check your answers by plugging them back into the original equation.

I hope this helps and encourages you to continue exploring the mysteries of logarithms. Good luck!