Question

log1664-log 6416 is equal to

Answer 1

your question is -> evaluate the value of $log(16)^{(64)}-log(64)^{(16)}$

we know from logarithmic properties

• log(m^n) = nlog(m)
• log(m) - log(n) = log(m/n)

first resolve log(16)^(64) and log(64)^(16) into simple form.

so, log(16)^(64) = log(2⁴)^(64)

= log(2)^{4 × 64}

= log(2)^(256)

= 256log(2)

similarly, log(64)^(16) = log(2^6)^(16)

= log(2)^(6 × 16)

= log(2)^(96)

= 96log(2)

now, log(16)^(64) - log(64)^(16)

= 256log(2) - 96log(2)

= 160log(2)

we know, log2 ≈ 0.301

so, 160log(2) ≈ 160 × 0.301 = 48.16

Answer 2

Answer:

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

Step-by-step explanation:

I think your question is

$log_{16}64-log_{64}16$

Take

$log_{16}64=x$

$\implies\:16^x=64$

$\implies\:(4^2)^x=4^3$

$\implies\:4^{2x}=4^3$

$\implies\:2x=3$

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

$\implies\:log_{16}64=\frac{3}{2}$

Using

$\boxed{log_ab=\frac{1}{log_ba}}$

$\implies\:log_{64}16=\frac{1}{log_{16}64}$

$\implies\:log_{64}16=\frac{2}{3}$

Now,

$log_{16}64-log_{64}16$

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

$=\frac{9-4}{6}$

$=\frac{5}{6}$

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