Logarithms can be naughty

Algebra Level 3

If $c = \log_{30}3$ and $d = \log_{30}5$ , then find the value of

$\large \log_{30}8$

2(1+c+d)

2(1-c-d)

2(1+c-d)

3(1+c-d)

3(1+c+d)

3(1-c-d)

1 solution

Vishwak Srinivasan
Jul 6, 2015

$\log_{30}3 + \log_{30}5 = \log_{30}15$

$\Rightarrow \log_{30}15 = \log_{30}\left(\dfrac{30}{2}\right) = \log_{30}30 - \log_{30}2 = 1 - \log_{30}2$

$\Rightarrow c + d = 1 - \log_{30}2$

$\log_{30}8 = 3\log_{30}2 = \boxed{3(1-c-d) }$

Great solution! Simple!

Ano Maly - 5 years, 11 months ago

Thanks. Cheers.

Vishwak Srinivasan - 5 years, 11 months ago

Good solution.

Sai Ram - 5 years, 9 months ago