If log 18 to the base 12=a and log 54 to the base 24=b,prove that ab+5(a-b)=1?

2 Answers
Jul 11, 2017

see explanation

Explanation:

Please check to make sure I've read the problem correctly. Logarithms may be written wrong...

  1. \log_12(18)=a
  2. \log_24(54)=b

ab+5(a-b)\stackrel{?}{=}1
(\log_12(18)\cdot\log_24(54))+5(\log_12(18)-\log_24(54))\stackrel{\color(olive)(\sqrt{})}{=}1

Jul 12, 2017

Please see below.

Explanation:

As log_12(18)=a, we have 12^a=18

i.e. (2^2xx3)^a=2xx3^2

or 2^(2a-1)xx3^(a-2)=1

similarly as log_24(54)=b, we have 24^b=54

i.e. (2^3xx3)^b=2xx3^3

or 2^(3b-1)xx3^(b-3)=1

Comparing the two 2a-1=3b-1 or 2a-3b=0 ........(A)

and a-2=b-3 or a-b+1=0 i.e. 2a-2b+2=0 ........(B)

Subtracting (A) from (B), we get

b+2=0 i.e. b=-2

and a=b-1=-3

hence ab+5(a-b)

=(-3)xx(-2)+5xx(-1)=6-5=1

(Note that (B) gives a-b=-1.)