# bloody indices

• September 11th 2005, 04:09 AM
thebizzle
bloody indices
i got this question 5^y=1 and i have to solve the equation, i know the that y=0 i jus dont no how to explain it pls help
• September 11th 2005, 08:39 AM
hoeltgman
Try this:
$5^y = 1 \Leftrightarrow y \cdot \log_5 (5) = \log_5 (1)$
$\text{In that case:} \log_5 (5) = 1\; \text{and} \log_5 (1) = 0$
• September 11th 2005, 09:45 AM
ticbol
Here is one way.

5^y = 1
Take the log of both sides,
y*log(5) = log(1)
Since log(1) = 0, then,
y = 0 / log(5)
y = 0
• September 12th 2005, 11:44 AM
thebizzle
thank u for ur replies but i am in yr 12 and have not started on logarithms yet
• September 12th 2005, 11:59 AM
hemza
ouch
Try this:

you know 5/5 = 1 and x^a / x^b = x^(a-b) so 5 / 5 = 1 gives 5^1 / 5^1 = 1 and then, 5^(1-1) = 5^0 so y=0.