Simplify: log 64 to the base 16 - log 16 is equal to
answer =5/6
Answers
Step-by-step explanation:
Expression \log_{16}64-\log_{64}16log
16
64−log
64
16
To find : Simplify the expression ?
Solution :
We solve the expression into parts,
Let \log_{16}64-\log_{64}16=x-ylog
16
64−log
64
16=x−y
So, x=\log_{16}64x=log
16
64
Applying logarithmic property, \log_b x=a\Rightarrow b^a=xlog
b
x=a⇒b
a
=x
16^x=6416
x
=64
(2^4)^x=2^6(2
4
)
x
=2
6
2^{4x}=2^62
4x
=2
6
As base are same power equates,
4x=64x=6
x=\frac{6}{4}x=
4
6
x=\frac{3}{2}x=
2
3
Similar approach for another term,
So, y=\log_{64}16y=log
64
16
Applying logarithmic property, \log_b x=a\Rightarrow b^a=xlog
b
x=a⇒b
a
=x
64^y=1664
y
=16
(4^3)^y=4^2(4
3
)
y
=4
2
4^{3y}=4^24
3y
=4
2
As base are same power equates,
3y=23y=2
y=\frac{2}{3}y=
3
2
Substitute the values back,
\log_{16}64-\log_{64}16=\frac{3}{2}-\frac{2}{3}log
16
64−log
64
16=
2
3
−
3
2
\log_{16}64-\log_{64}16=\frac{9-4}{6}log
16
64−log
64
16=
6
9−4
\log_{16}64-\log_{64}16=\frac{5}{6}log
16
64−log
64
16=
6
5
Therefore, \log_{16}64-\log_{64}16=\frac{5}{6}log
16
64−log
64
16= 5/6