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## pottersheep Group Title Please help! Show that each statement is true (LOGS) [1/log base 5 of a] + [1/log base 3 of a] = [1/logbase 15a] one year ago one year ago

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1. pottersheep Group Title

sorry last one is log base 15 of a

2. satellite73 Group Title

lets add on the left

3. satellite73 Group Title

$\frac{\log_3(a)+\log_5(a)}{\log_3(a)\times \log_5(a)}$

4. satellite73 Group Title

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

5. satellite73 Group Title

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 $$\log_{15}(x)$$

6. satellite73 Group Title

do you know the change of base formula?

7. satellite73 Group Title

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

8. satellite73 Group Title

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

9. satellite73 Group Title

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 $\log_{15}(a)\times \log_{15}(3\times 5)=\log_{15}(a)\times 1$

10. satellite73 Group Title

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

11. satellite73 Group Title

cancel and you get what you want

12. pottersheep Group Title

oooooo Thanks!