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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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 $ $
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
thank u for ur replies but i am in yr 12 and have not started on logarithms yet
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 ?
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