I haven't calculus in a while. So can someone please help me with this, its for rates:
log (base 1/4) 0.25 = m
I know that m = 1, but how would I solve for it using the log method.
Isn't the log(base x) X = 1 ??
log (base 10) 10 = 1
log (base e) e = 1
so log(base 1/4) 1/4 = 1 ?? Check my thinking.
Yes I know it will be m=1. But what if I have something like
log (base 1/4) 0.33 = m
Then how would I solve something like that.
.m raised to the .25 power or the fourth root of m = 33 is what?
To solve the equation log(base 1/4) 0.25 = m, we can apply the definition of logarithms. The logarithm gives you the exponent to which the base must be raised to obtain the given value.
In this case, the base is 1/4, and the value is 0.25. Therefore, we can rewrite the equation as:
(1/4)^m = 0.25
Now, let's proceed with solving for m step by step:
Step 1: Convert the base to a different format.
Since 1/4 can be expressed as 4^(-1), we rewrite the equation as:
(4^(-1))^m = 0.25
Step 2: Apply the property of exponentiation.
When raising a power to another power, we multiply the exponents:
4^(-m) = 0.25
Step 3: Rewrite 0.25 as a fraction with a power of 4.
Since 0.25 is 1/4, we can represent it as 4^(-2) and rewrite the equation as:
4^(-m) = 4^(-2)
Step 4: Equate the exponents.
Since the bases are the same on both sides of the equation, we can equate the exponents:
-m = -2
Step 5: Solve for m.
To isolate the variable, multiply both sides of the equation by -1:
m = 2
Therefore, the solution to the equation log(base 1/4) 0.25 = m is m = 2.