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

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- Sep 11th 2005, 04:09 AMthebizzlebloody 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

- Sep 11th 2005, 08:39 AMhoeltgman
Try this:

$\displaystyle $5^y = 1 \Leftrightarrow y \cdot \log_5 (5) = \log_5 (1)$ $

$\displaystyle $\text{In that case:} \log_5 (5) = 1\; \text{and} \log_5 (1) = 0 $ $ - Sep 11th 2005, 09:45 AMticbol
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 - Sep 12th 2005, 11:44 AMthebizzle
thank u for ur replies but i am in yr 12 and have not started on logarithms yet

- Sep 12th 2005, 11:59 AMhemzaouch
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.

How about that ?