1. ## logs

Let y = loga(10) + loga(12) - loga(6)

Find a if y=1
Find a if y=-1
Find a if y=2

If you can show me any of the obove please let me know.

THANKS

2. Originally Posted by Mr_Green
Let y = loga(10) + loga(12) - loga(6)

Find a if y=1
Find a if y=-1
Find a if y=2

If you can show me any of the obove please let me know.

THANKS
log(10)/log(a) + log(12)/log(a) - log(6)/log(a) = log(20)/log(a)

1 = log(20)/log(a)

log(a) = log(20)

ln of both sides; a = 20

You do them now; answers, for:

y = -1, you should get: a = 1/20

y = 2, you should get: a = 2*sqrt(5)

3. Originally Posted by Mr_Green
Let y = loga(10) + loga(12) - loga(6)

Find a if y=1
Find a if y=-1
Find a if y=2
Combine the logarithms,
$\displaystyle y=\log_a (10 \cdot 12/6)=\log_a 20$
Thus, by definition of logarithms,
$\displaystyle a^y=20$ (since $\displaystyle a>0$)
Thus, (if $\displaystyle y\not = 0$)
$\displaystyle a=20^{1/y}$
So just substitute the value of "y" to take the value of "a"