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.