Question

Log25 log32 log6 36

Answer 1

$log_{25}log_{32}log_{6}36=log_{5^{2}}log_{2^{5}}log_{6}6^2= \frac{1}{2} log_{5} \frac{1}{5} log_{2}2log_{6}6=\frac{1}{2} log_{5} \frac{1}{5} log_{2}2$
$=\frac{1}{2} log_{5}5^{-1}=-\frac{1}{2} log_{5}5=-\frac{1}{2}=-0,5$

Answer 2

To solve the expression log25 + log32 + log6 + 36, we need to use the properties of logarithms.

First, let's simplify the expression using the properties of logarithms:

log25 + log32 + log6 + 36

Using the rule of logarithmic addition, we can combine the logarithms with the same base:

log25 + log32 + log6 = log(25 * 32 * 6)

Now, let's calculate the value inside the logarithm:

25 * 32 * 6 = 4800

So, the simplified expression becomes:

log4800 + 36

Now, we can evaluate the logarithm using the base 10 logarithm (log10):

log4800 ≈ 3.6812

Finally, we can add this value to 36:

3.6812 + 36 ≈ 39.6812

Therefore, the value of the expression log25 + log32 + log6 + 36 is approximately 39.6812.