intersections of a logarithmic and exponential function

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fishspawned

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Feb 12, 2015
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Hi there,

I have become curious about how to approach this problem:
We know that an exponential function and its inverse [log function] intersects along the y=x axis
However I have been having difficulty figuring out how to calculate that point.

Here is an example, and i will show how far i get:

consider (1/5)^x=y and log[base 1/5]x=y
so if they intersect at y=x i took the first equation
and let x=y...

(1/5)^x=x
log(1/5)^x= logx
log(1/5)=(logx)/x

and here i am stumped. that expression on the right gets me.

any help would be greatly appreciated and thank you in advance
 
fishspawned said:
Hi there,

I have become curious about how to approach this problem:
We know that an exponential function and its inverse [log function] intersects along the y=x axis
Click to expand...
It is good, well, excellent, that you are curious. But your statement is not exactly true. The two functions are symmetrical along the line y=x but may NOT intersect. Just consider y=lnx and y=x. They do not intersect! If they did intersect at say x=3, then yes, ln(3) would equal e^3.
 
Last edited:
Jomo said:
It is good, well excellent, that you are curious. But your statement is not exactly true. The two functions are symmetrical along the line y=x but may NOT intersect. Just consider y=lnx and y=x. They do not intersect! If they did intersect at say x=3, then yes, ln(3) would equal e^3.
Click to expand...

Got curious, so worked it out. If the lines intersect then
ax = x
or
a = x1/x
If we analyze the function
a(x) = x1/x
we note that it has a global maximum at x = e. Thus, if
ax = x
and a and x are real positive numbers, a is less than or equal to e1/e ~ 1.445
 
Thank you Jomo and Ishuda,
It is true - and I should have not stated that all exponential functions and their inverses will intersect at y=x. Quite obviously, I must first assume that the exponential function actually intersects y=x at some point, and only certain bases will allow this.

But this still does does not give me a solid way to solve the equation that I originally was looking at.

which now can be looked at in two ways:
log(1/5)=(logx)/x
or
x^(1/x) = 1/5

in this particular case, I know it will intersect as base 1/5 is good enough to do so, but how to solve?
i'm basically trying to figure out the xth square root of a number x that is 1/5... or (1/5)^x = x....
 
fishspawned said:
Thank you Jomo and Ishuda,
It is true - and I should have not stated that all exponential functions and their inverses will intersect at y=x. Quite obviously, I must first assume that the exponential function actually intersects y=x at some point, and only certain bases will allow this.

But this still does does not give me a solid way to solve the equation that I originally was looking at.

which now can be looked at in two ways:
log(1/5)=(logx)/x
or
x^(1/x) = 1/5

in this particular case, I know it will intersect as base 1/5 is good enough to do so, but how to solve?
i'm basically trying to figure out the xth square root of a number x that is 1/5... or (1/5)^x = x....
Click to expand...
There is no straight up way to find the answer that I know of. You will have to use 'a guess and correct' method. x=0.47 is close for x=0.2x I believe. If that's not close enough try 0.20.47 and keep going if that's not close enough.
 
fishspawned said:
Thank you Jomo and Ishuda,
It is true - and I should have not stated that all exponential functions and their inverses will intersect at y=x. Quite obviously, I must first assume that the exponential function actually intersects y=x at some point, and only certain bases will allow this.
Click to expand...
Nope, you can use any base. If you function does not intersect the line y=x, then simple shift the function vertically and/or horizontally and it will intersect y=x. AND the base will not change.
 
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