1. ## Logarithms

Hi! I understand the concepts regarding logarithms, but I'm having some trouble showing that the following statement is true, if you could help me out I would really appreciate it 3/(log[2]a) - 2/(log[4]a) = 1/(log[0.5]a) *the square brackets indicate that the log is in the base of the number inside the brackets *THEY ARE NOT NATURAL LOGARITHMS* Thanks again!

2. Originally Posted by teddy7
Hi! I understand the concepts regarding logarithms, but I'm having some trouble showing that the following statement is true, if you could help me out I would really appreciate it 3/(log[2]a) - 2/(log[4]a) = 1/(log[0.5]a) *the square brackets indicate that the log is in the base of the number inside the brackets *THEY ARE NOT NATURAL LOGARITHMS* Thanks again!
Hint:

Assume Log2(a) = x → 2x = a

calculate Log4(a) and Log0.5(a) as a function of 'x'

3. Hello, teddy7!

Formula: .$\log_b(a) \:=\:\dfrac{\ln a}{\ln b}$

$\displaystyle\text{Show that: }\:\frac{3}{\log_2(a)} - \frac{2}{\log_4(a)} \;=\;\frac{1}{\log_{0.5}(a)}$

The left side is: .$\dfrac{3}{\frac{\ln(a)}{\ln(2)}} - \dfrac{2}{\frac{\ln(a)}{\ln(4)}} \;\;=\;\; \dfrac{3\ln(2)}{\ln(a)} - \dfrac{2\ln(4)}{\ln(a)} \;\;=\;\;\dfrac{\ln(2^3)}{\ln(a)} - \dfrac{\ln(4^2)}{\ln(a)}$

. . . . . . . . . $=\;\;\dfrac{\ln(8)}{\ln(a)} - \dfrac{\ln(16)}{\ln(a)} \;\;=\;\; \dfrac{\ln(8) - \ln(16)}{\ln(a)} \;\;=\;\;\dfrac{\ln\left(\frac{8}{16}\right)}{\ln( a)}$

. . . . . . . . . $=\;\;\dfrac{\ln(0.5)}{\ln(a)} \;=\;\dfrac{1}{\frac{\ln(a)}{\ln(0.5)}} \;\;=\;\;\dfrac{1}{\log_{0.5}(a)}$

4. Originally Posted by teddy7
Hi! I understand the concepts regarding logarithms, but I'm having some trouble showing that the following statement is true, if you could help me out I would really appreciate it 3/(log[2]a) - 2/(log[4]a) = 1/(log[0.5]a) *the square brackets indicate that the log is in the base of the number inside the brackets *THEY ARE NOT NATURAL LOGARITHMS* Thanks again!
Another way:

Log2(a) = x → 2x = a → 4x/2 = a → Log4(a) = x/2

Log2(a) = x → 2x = a → (1/2)-x = a → Log1/2(a) = -x

Then

3/Log2(a) - 2/Log4(a)

= 3/x - 2/(x/2)

= 3/x - 4/x

= 1/(-x)

= 1/Log1/2(a)

5. Regarding
Originally Posted by soroban
. . . . . . . . . $\ \ ... \ =\;\;\dfrac{\ln(0.5)}{\ln(a)} \;=\;\dfrac{1}{\frac{\ln(a)}{\ln(0.5)}} \;\;=\;\;\dfrac{1}{\log_{0.5}(a)} \ \$
and
Originally Posted by Subhotosh Khan
... = 1/Log1/2(a)
$\ \ \$ the answer's form can also be streamlined to avoid the fractional form as:

$\dfrac{\ln(0.5)}{\ln(a)} \ =$

$\dfrac{\ln(2^{-1})}{\ln(a)} \ =$

$\dfrac{- \ln(2)}{\ln(a)} \ =$

$\boxed{- \log_a(2)}$

Edit: Yes, it is the "required form," but it is a useful thing to know, especially if the OP were to be shown an alternate form in the future.

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