Question on logs???

Question

log base 2 of 56 minus log base 4 of 49. i know the answer is 3 but i need to show my work. please help!

Answer 1

To perform operations on logs, they must have the same base. Because 4=2², log(4)49=½log(2)49
=log(2)49^(½)
=log(2)sqrt(49)
=log(2)7

Now, log(2)56-log(2)7 = log(2)56/7
=log(2)8=3

Answer 2

Let x = log[base 2](56), and
Let y = log[base 4](49)

We can use the change of base formula on log[base 4](49). Remember that
log[base a](c) = log[base b](c) / log[base b](a)

Let's change this to base 2.

log[base 4](49) = log[base 2](49) / log[base 2](4)

But log[base 2](4) = 2, so we have

log[base 4](49) = (1/2) log[base 2](49)

We can move the (1/2) as a power of 49 as per the log property
c log[base b](a) = log[base b](a^c).

log[base 4](49) = log[base 2](49^(1/2))

49^(1/2) is equal to 7, so we have

log[base 2](7)

Remember that originally we have

log[base 2](56) - log[base 4](49)

and we just discovered that log[base 4](49) = log[base 2](7), so

log[base 2](56) - log[base 2](7)

Now we can combine these two logs as per the log property
log[base 2](a) - log[base 2](c) = log[base 2](a/c), giving us

log[base 2](56/7)
log[base 2](8)

This is asking, "2 to the what power is equal to 8?" to which the answer is 3.

Answer 3

u can solve it -
log 56 - (log 49)/(log 4) 1. ALL base 2

becauz log c (base a) = log c (base b)/log b (base a)

so, 1. =
log 56 - (log 49)/2
= (2 log 56 - log 49)/2
= (log (56^2) - log 49)/2
= log(56^2 / 49)/2
= (log 64) /2
= 6/2
=3
all done by log rules, u can search for them