Question

Please help! Show that each statement is true (LOGS)

[1/log base 5 of a] + [1/log base 3 of a] = [1/logbase 15a]

Answer 1

sorry last one is log base 15 of a

Answer 2

lets add on the left

Answer 3

\[\frac{\log_3(a)+\log_5(a)}{\log_3(a)\times \log_5(a)}\]

Answer 4

now lets flip it and see if we can show it is the same as \(\log_{15}(a)\)

Answer 5

that is, can we show
\[\frac{\log_3(a)\times \log_5(a)}{\log_3(a)+\log_5(a)}=\log_15(a)\] and i think we can do it using the change of base formula and writing everything in terms of \(\log_{15}(x)\)

Answer 6

do you know the change of base formula?

Answer 7

you get a nasty compound fraction on the left, i will see if i can write it
\[\frac{\frac{\log_{15}(a)}{\log_{15}(5)}\times \frac{\log_{15}(a)}{\log_{15}(5)}}{\frac{\log_{15}(a)}{\log_{15}(5)}+\frac{\log_{15}(a)}{\log_{15}(3)}}\]

Answer 8

to clear the compound fraction, multiply top and bottom by \(\log_{15}(3)\log_{15}(5)\)

Answer 9

the numerator will be
\[\log_{15}(a)\times \log_{15}(a)\] and the denominator will be
\[\log_{15}(a)\log_{15}(5)+\log_{15}(a)+\log_{15}(5)\] factor as
\[\log_{15}(a)\left(\log_{15}(5)+\log_{15}(3)\right)\] which gives
\[\log_{15}(a)\times \log_{15}(3\times 5)=\log_{15}(a)\times 1\]

Answer 10

typo above, second line should be
\[\log_{15}(a)\log_{15}(5)+\log_{15}(a)\log_{15}(5)\]

Answer 11

cancel and you get what you want

Answer 12

oooooo Thanks!