Solving A Nice Logarithmic System

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[mikecaetano]

The graph looks like the infield of a baseball park.

[MichaelRothwell]

Thank you for this beautiful home-made problem and solution. The graph of the relations is amazing!

[GirishManjunathMusic]

Given:
x↑log(y) + y↑log(x) = 2
x↑log(x) + y↑log(y) = 11
To find:
x and y:

Rewriting a as 10↑(log(a)):
(10↑log(x))↑log(y) + (10↑log(y))↑log(x) = 2

Rewriting (10↑a)↑b = 10↑ab:
10↑log(x)·log(y) + 10↑log(y)·log(x) = 2
2·(10↑log(x)·log(y)) = 2
10↑log(x)·log(y) = 1

As 1 = 10↑0:
10↑log(x)·log(y) = 10↑0

Comparing exponents:
log(x)·log(y) = 0

Either log(x) = 0, or log(y) = 0, or log(x) = log(y) = 0.
If log(x) = 0:
x↑0 + y↑(log(y) = 11
y↑(log(y) = 10

Rewriting a = 10↑log(a):
(10↑log(y))↑log(y) = 10
10↑(log(y))² = 10

Comparing exponents:
(log(y))² = 1
log(y) = ± 1
y = 10 or 0.1.

If log(y) = 0, then the inverse of the previous process results in x = 10 or 0.1.

If log(x) = log(y) the equations are not satisfied, as they imply 11 = 2.

Thus, (x, y) can be (1, 10), (1, 0.1), (10, 1), (0.1, 1).

All 4 of these work for both equations by cross-checking.

[BlaqRaq]

I’m appreciative of graphing computers because you get actual visual values very quickly. One of my fetish was sketching the curve for all types of functions with holes and asymptotes. I loved maths. Keep doing what you’re doing. Great job!

[angelmendez-rivera]

The equations to solve are x^log(10, y) + y^log(10, x) = 2 & x^log(10, x) + y^log(10, y) = 11 for real x, y > 0, so that log(10, x) and log(10, y) are well-defined. Let x = 10^a for arbitrary real a, and y = 10^b for arbitrary real b, hence 10^(a·b) + 10^(a·b) = 2 & 10^(a^2) + 10^(b^2) = 11. 10^(a·b) + 10^(a·b) = 2 is equivalent to 10^(a·b) = 1, which is equivalent to a·b = 0, which is equivalent to a = 0 or b = 0, since the real numbers have no nontrivial zero divisors. 10^(a^2) + 10^(b^2) = 11 & [a = 0 or b = 0] is equivalent to [a = 0 & 10^(a^2) + 10^(b^2) = 11] or [10^(a^2) + 10^(b^2) = 11 & b = 0]. Therefore, [a = 0 & b^2 = 1] or [a^2 = 1 & b = 0]. This means (x, y) = (1, 1/10) or (x, y) = (1, 10) or (x, y) = (1/10, 1) or (x, y) = (10, 1). In fact, since the order does not matter, you could even state this more succinctly as {x, y} = {1/10, 1} or {x, y} = {1, 10}.

[armacham]

Since you know x > 0, you can substitute x = 10^a, where a is any real number
And you can do the same thing with y = 10^b
this gets rid of the logarithms, both equations are just in terms of a and b, which makes it much easier to solve

[StuartSimon]

What would be more interesting to me about the graph of the second equation would be to see if I could find where the horizontal and vertical tangents are as well as those with slopes 1 and -1.

[tryingtomakeanamebelike]

logx=a
logy=b
2 x 10^ab= 2
ab=0
10^a^2 + 10^b^2 = 11
a and b both being 0 doesn’t work, so substitute one variable = 0 into second equation
x=10, 1/10, y=1
x=1, y=10, 1/10

[qwang]

As 10^log(v) = v, the first equation is actually: 10^[log(x)*log(y)] + 10^[log(x)*log(y)] = 2. So 10^[log(y)*log(x)] = 1. log(x)*log(y) = 1, log(x) = 0 or log(y) = 0,
x = 1 or y = 1. If x = 1, then from the 2nd equation: y^(logy) = 10. (log(y))^2 = 1. log(y) = 1 or -1, y = 10 or 1/10. Similarly, if y = 1, then x = 10 or 1/10. So 4 solutions.
...

[billtruttschel]

When you type these out, you should put a space between the log and the argument.

[yoav]

Nice problem, great explanation, nice solution and interesting graph at the end.what do we need more?😃💯

[roshanramesh]

Doing this in my mind boosts your confidence like anything. Different method though. Substitute each term and convert it into 10^ whatever and solve.

[MarcelCox]

I solved the problem for natural logarithm which is not more difficult than the decimal one, except that you don't get such a neat reply.

[mustafasaracaloglu]

what about z^2 - 11z +1 =0 where z = (logx)^2 and comes from (logy)^2 = 1/(logx)^2 .

[scottleung]

I got (10, 1) from guessing and checking.

[rihamnour]

Hi, thanks for the helpful videos and do you have a page on Facebook because I want to ask you questions about differential equations