What is 3log_{2}(3)+5log_{2}(1)+4log_{2}(b)-5 ?

The solution to 3log_{2}(3)+5log_{2}(1)+4log_{2}(b)-5 is log_{2}(27b^4)-5

Popular >

simplify 3log_{2}(3)+5log_{2}(1)+4log_{2}(b)-5

simplify

3 lo g_{2} (3) + 5 lo g_{2} (1) + 4 lo g_{2} (b) - 5

lo g_{2} (27 b^{4}) - 5

Solution steps

3 lo g_{2} (3) + 5 lo g_{2} (1) + 4 lo g_{2} (b) - 5

5 lo g_{2} (1) = 0

= 3 lo g_{2} (3) + 0 + 4 lo g_{2} (b) - 5

Group like terms

= 3 lo g_{2} (3) + 4 lo g_{2} (b) + 0 - 5

Subtract the numbers:

0 - 5 = - 5

= 3 lo g_{2} (3) + 4 lo g_{2} (b) - 5

3 lo g_{2} (3) = lo g_{2} (3^{3})

4 lo g_{2} (b) = lo g_{2} (b^{4})

= lo g_{2} (3^{3}) + lo g_{2} (b^{4}) - 5

Simplify

lo g_{2} (3^{3}) + lo g_{2} (b^{4}) : lo g_{2} (27 b^{4})

= lo g_{2} (27 b^{4}) - 5

13.375 ln (\frac{1390}{1390 - 1250})

2 lo g_{2} (x) - 4 lo g_{2} (y) - 5 lo g_{2} (z)

lo g_{10} (\frac{x ^{2} + 9}{x + 1})

ln (\frac{1}{0.32})

lo g_{4} (\frac{f}{16})

What is 3log_{2}(3)+5log_{2}(1)+4log_{2}(b)-5 ?
The solution to 3log_{2}(3)+5log_{2}(1)+4log_{2}(b)-5 is log_{2}(27b^4)-5