logs help please..?

G

Guest

Guest
How would i solve this on paper (no calculator)

Log5(x)+3=y i need to fing the x-intercept
 
X intercept is when y = 0:

\(\displaystyle Log _5 x + 3 = 0\)

\(\displaystyle Log _5 x = -3\)

\(\displaystyle 5^{Log _5 x} = 5^{-3}\)

\(\displaystyle x = \frac{1}{5^3} = \frac{1}{125}\)
 
thank you

well that seem obvios sorta. so you multiplied by 5 to cancel the log? my math teacher seemed to miss explaing that. :shock:
 
Re: thank you

adon said:
well that seem obvios sorta. so you multiplied by 5 to cancel the log? my math teacher seemed to miss explaing that. :shock:
Click to expand...
No, I raised both sides to the power of 5.

Remember:

\(\displaystyle \L a^{Log_a x} = x\)
 
I dont know what that means but thank you for trying. i started logs today ( all i know is y=log(a)X is that same as a^y=x)
 
adon said:
I dont know what that means but thank you for trying. i started logs today ( all i know is y=log(a)X is that same as a^y=x)
Click to expand...

In simple terms, if raise 5 to Log<sub>5</sub>63, you will get 63. To cancel a Log, raise the whole expression to whatever the base of the Log is.

Using your expression, Log<sub>5</sub>x = -3 implies that 5<sup>-3</sup> = x. Good?
 
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