Question

Найдите log25 (162), если log5 (2)=a, log3 (5)=b

Answer 1

$log_52=a$

$log_35=b$

$log_{25}162=\frac{log_5162}{log_525}=\frac{log_5(2\cdot81)}{log_55^2}=$

$\frac{log_5(2\cdot3^4)}{2log_55}=\frac{log_52+log_53^4}{2}=$

$\frac{a+4log_53}{2}=\frac{a+4 \cdot \frac{1}{log_35} }{2}=$

$\frac{a+\frac{4}{b} }{2}=\frac{\frac{ab+4}{b}}{2}=\frac{ab+4}{2b}$

Answer 2

To find the value of log25(162), we can use the change of base formula for logarithms. The change of base formula states that logₐ(b) = logₓ(b) / logₓ(a), where x can be any base.

Given that log₅(2) = a and log₃(5) = b, let's express log25(162) in terms of a and b:

log25(162) = log₅(162) / log₅(25)

To express log₅(162) in terms of a and b, we need to find a relation between 162 and 5 using the given logarithmic identities.

Since log₅(2) = a, we can rewrite it as:

5^a = 2

Similarly, since log₃(5) = b, we can rewrite it as:

3^b = 5

Now, let's find the relation between 162 and 5:

162 = 2 * 3^4

Using the above relation, we can rewrite log₅(162) as:

log₅(162) = log₅(2 * 3^4)

Applying logarithmic properties, we can split it into two logarithms:

log₅(162) = log₅(2) + log₅(3^4)

Since log₅(2) = a, we have:

log₅(162) = a + log₅(3^4)

Since 3^4 = (3^2)^2 = 9^2, we can rewrite it as:

log₅(162) = a + log₅(9^2)

Using the property logₐ(b^c) = c * logₐ(b), we can rewrite it as:

log₅(162) = a + 2 * log₅(9)

Since 9 = 3^2, we can rewrite it further:

log₅(162) = a + 2 * log₅(3^2)

Using logarithmic properties, we can bring the exponent out as a coefficient:

log₅(162) = a + 2 * 2 * log₅(3)

Since log₃(5) = b, we can replace log₅(3) with b:

log₅(162) = a + 2 * 2 * b

Simplifying further, we have:

log₅(162) = a + 4b

Now, let's substitute the values of a and b:

log₅(162) = a + 4b log₅(162) = a + 4 * log₃(5)

Since log₃(5) = b, we can rewrite it as:

log₅(162) = a + 4b log₅(162) = a + 4 * b

Therefore, log25(162) can be expressed as a + 4b.