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log is the inverse of exponents; and exponents are the inverse of logs. What is your question regarding?
i dont understand how to do them at all. how would you graph y=3log(base5)x. how would you go about that
Hey amistre can you please help me with 2 antiderivative questions after you are done here?
Since y = 3 log5(x) is the same as: y = log5(x^3) we can more easily graph its inverse and then "flip" the graph about the y=x line.
you got a question posted we can go to :)
did you get the second equation from switching the x and ys and solving for y?
did you see my question
yeah, but I messed it up in my head the first time :)
meet me there when you can help me
mini: log graphs can be hard to do without a calulator; so we can rewrite it to a more familiar form... do you agree?
but how do you make it in the other form?
y = 3 log5(x) ; divide by 3 y/3 = log5(x) ; take the 5^ of each side 5^(y/3) = 5^(log5(x)) ; 5^(log5) cancel each other out. 5^(y/3) = x Do you agree? Are you familiar with the rules for logs?
i understand how you did that. when is that that you go about switching the x and y to solve?
When it makes the graphing easier you can modify it. All you are doing is solving for x instead of y, so keep aware of that
Would you agree that 5^(y/3) is easier to plot for and solve than log5(x) ? :)
how do you do for example logbase8 4096=4
Do you mean: log8(4096) = 4 ??
is it just 8^4=4096
that is what is known as an identity. one side equals the other. Lets take for example: logB(x) = y this means that B^y=x We can take your equation for instance: log8(4096) = 4 means: 8^4 = 4096, we can test that by either pen and paper , or calculator :)
35^log35 = 1 and we are left with "x"
can you do the inverse of y=log1/4 x out step by step please?
is that log base (1/4)?
i understand that one now but how do you do the second one?
y = ln(6x) correct?
just y= ln 6x not base 6x
"ln" is just a special way they write log to the base "e"
y = ln(6x) e^y = e^ln(6x) e^y = 6x (e^y)/6 = x
how about y= ln (x+2)
e^y = x+2 e^y -2 = x
so for the graph of y=log8 x-2 you would do 8^y=x-2 and then fill in values for y such as 0 which would be 1=x-2 x=3?