Question

Simplify: log 64 to the base 16 - log 16 to the base 64

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\log_{16}64-\log_{64}16=\frac{5}{6}$

Step-by-step explanation:

Given : Expression $\log_{16}64-\log_{64}16$

To find : Simplify the expression ?

Solution :

We solve the expression into parts,

Let $\log_{16}64-\log_{64}16=x-y$

So, $x=\log_{16}64$

Applying logarithmic property, $\log_b x=a\Rightarrow b^a=x$

$16^x=64$

$(2^4)^x=2^6$

$2^{4x}=2^6$

As base are same power equates,

$4x=6$

$x=\frac{6}{4}$

$x=\frac{3}{2}$

Similar approach for another term,

So, $y=\log_{64}16$

Applying logarithmic property, $\log_b x=a\Rightarrow b^a=x$

$64^y=16$

$(4^3)^y=4^2$

$4^{3y}=4^2$

As base are same power equates,

$3y=2$

$y=\frac{2}{3}$

Substitute the values back,

$\log_{16}64-\log_{64}16=\frac{3}{2}-\frac{2}{3}$

$\log_{16}64-\log_{64}16=\frac{9-4}{6}$

$\log_{16}64-\log_{64}16=\frac{5}{6}$

Therefore, $\log_{16}64-\log_{64}16=\frac{5}{6}$