Prove:
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))
since 4 = 2^2, log_4(a) = 1/2 log_2(a)
since 1/2 = 2^-1, log_(1/2)(a) = -log_2(a)
so, if we let x = log_2(a), we have
3/x - 2/(x/2) = 1/(-x)
3/x - 4/x = -1/x
(4-3)/x = -1/x
1/x = -1/x
???
Is there a typo somewhere ?
i.imgur[dot]com/hE0sWBt[dot]gif
Make sure its just like that, it has to be solvable o.o
er "prove-able"
To prove the given equation:
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))
We'll start by simplifying each term using logarithmic identities.
Let's first simplify the left side of the equation:
3/(log_2 (a)) - 2/(log_4 (a))
Using the change of base formula for logarithms, we can rewrite log_4 (a) as log_2 (a) / log_2 (4).
3/(log_2 (a)) - 2/(log_2 (a) / log_2 (4))
Next, we'll simplify the denominator by evaluating log_2 (4). Since 2^2 = 4, log_2 (4) = 2.
3/(log_2 (a)) - 2/(log_2 (a) / 2)
This simplifies to:
3/(log_2 (a)) - 4/(log_2 (a))
Now, we have a common denominator (log_2 (a)) and we can combine the fractions:
(3 - 4) / (log_2 (a))
Now, let's simplify the right side of the equation:
1/(log_(1/2)(a))
Using the change of base formula again, we can rewrite this as:
1 / (log_2 (a) / log_2 (1/2))
Since 2^(-1) = 1/2, log_2 (1/2) = -1.
1 / (log_2 (a) / -1)
Now, dividing by a fraction is the same as multiplying by its reciprocal. Therefore, we can rewrite this as:
1 * (-1 / log_2 (a))
Now, simplifying:
-1 / (log_2 (a))
Comparing the simplified left side and right side of the equation, we can see that they are the same:
(3 - 4) / (log_2 (a)) = -1 / (log_2 (a))
Since both sides of the equation simplify to the same value, we have proved that:
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a)) is true for any value of 'a'.
Hence, the equation is proven.