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

Question

35878 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-11T18:13:06

see explanation

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

#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#

Answer 2 · 2017-07-12T09:15:30

Please see below.

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#.)