Write the following as a single logarithm. Assume all variables are positive. $$ 2\left(\log _{2}(5)+\log _{2}(b)\right)-\log _{2}(6)= $$ The answer format in lowercase characters is: log_base (number) Spaces in the answer are optional.

Question

UpStudy AI · Accepted Answer

Given the expression $$2\left(\log _{2}(5)+\log _{2}(b)\right)-\log _{2}(6)$$, we can simplify it by using the properties of logarithms.

Using the property $$\log _{a}(b) + \log _{a}(c) = \log _{a}(bc)$$, we can rewrite the expression as:
$$2\left(\log _{2}(5)+\log _{2}(b)\right)-\log _{2}(6) = 2\log _{2}(5b) - \log _{2}(6)$$

Now, using the property $$a\log _{b}(x) = \log _{b}(x^a)$$, we can further simplify the expression:
$$2\log _{2}(5b) - \log _{2}(6) = \log _{2}((5b)^2) - \log _{2}(6)$$

Using the property $$\log _{a}(b) - \log _{a}(c) = \log _{a}\left(\frac{b}{c}\right)$$, we can rewrite the expression as:
$$\log _{2}((5b)^2) - \log _{2}(6) = \log _{2}\left(\frac{(5b)^2}{6}\right)$$

Therefore, the expression $$2\left(\log _{2}(5)+\log _{2}(b)\right)-\log _{2}(6)$$ can be written as a single logarithm:
$$\log _{2}\left(\frac{(5b)^2}{6}\right)$$

So, the answer in lowercase characters is: log_base (number) = log_base (5b^2/6)
log₂(25b²/6)

Write the following as a single logarithm. Assume all variables are positive. \( 2\left(\log _{2}(5)+\log _{2}(b)\right)-\log _{2}(6)= \) The answer format in lowercase characters is: log_base (number) Spaces in the answer are optional.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions