How do you write log_6 5 as a logarithm of base 4?

Apr 18, 2016

${\log}_{6} 5 = 0.7737 \times {\log}_{4} 5$

Explanation:

Let ${\log}_{x} a = p$ and ${\log}_{c} x = q$.

i.e. ${x}^{p} = a$ and ${c}^{q} = x$ and hence $a = {\left({c}^{q}\right)}^{p} = {c}^{p q}$

i.e. ${\log}_{c} a = p \times q$ or ${\log}_{c} a = {\log}_{x} a \times {\log}_{c} x$-------(A)

Hence ${\log}_{6} 5 = {\log}_{4} 5 \times {\log}_{6} 4$...........(B)

(A) also tells us that ${\log}_{x} a = {\log}_{c} \frac{a}{\log} _ c x$

and hence ${\log}_{6} 4 = {\log}_{10} \frac{4}{\log} _ \left(10\right) 6$ and putting this in (B)

${\log}_{6} 5 = {\log}_{4} 5 \times \log \frac{4}{\log} 6 = \frac{0.6021}{0.7782} \times {\log}_{4} 5 = 0.7737 \times {\log}_{4} 5$

Apr 18, 2016

Use the change of base formula or solve an equation using log base 4.

Explanation:

Change of base formula

${\log}_{b} x = {\log}_{c} \frac{x}{\log} _ c b$

So ${\log}_{6} 5 = {\log}_{4} \frac{5}{\log} _ 4 6$

Solve an equation

If you don't remember the change of base formula (or if you want to see where it comes from)

Let $x = {\log}_{6} 5$

So ${6}^{x} = 5$

Because we want log base 4, take that log on both sides:

${\log}_{4} \left({6}^{x}\right) = {\log}_{4} 5$

Now use the exponent property of logarithms:

$x {\log}_{4} 6 = {\log}_{4} 5$.

Finally divide by ${\log}_{4} 6$ to get

$x = {\log}_{4} \frac{5}{\log} _ 4 6$.