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Solve,
3^(log 27 base x) - 2^(log 20 base y) = 7
x^y + y^x = 17
x and y is integer. I need the steps. thank you

This looks like maths homework Gan :-)

I wont give you the answer you need to do it I will give you the steps

STEP 1: Get equation 1 on a common base log

Here is a reference for you if you get stuck (http://www.mathstat.strath.ac.uk/basicmaths/233_lawsoflogarithms.html)
Look at law (vi)

Log c base a = log c base b / log a base b

note**** => I have avoided using x to not confuse you

Choose either base x or base y but one of them has to go so you have a common base log


STEP 2: Now you have eq 1 on a common base look for log proof exchanges

log(a/b) = log a - log b

This works for any base

So combine and simplify equation 1 on its common base log


STEP 3: Run antilogs to both sides to remove the logs

STEP 4: Now solve the 2 standard simultaneous equations

No No No...
If I log both sides on equation 1,
log (3^(log 27 base x) - 2^(log 20 base y)) = log(7)

You cannot seperate RHS. How to simplify???

Did you read the steps ... in above you didnt first put them on a common base log and second you need to antilog not log both sides but only after putting on a common base.

SO STEP 1:

You need to get them on a common log base

EQN 1: 3^(log 27 base x) - 2^(log 20 base y) = 7

So lets say you want to get rid of the "log 20 base y" term and convert to base x

I gave you the base conversion formula

Log c base a = log c base b / log a base b


So: Log 20 base y = log 20 base x / log y base x


Substitue the above into your formula


3^(log 27 base x) - 2^(log 20 base x / log y base x) = 7


Now you only have base x logs so simplify it

then just continue.... I wish to see ur steps

Gan this is far too long mathematics for a physics forum and I am thinking you are not getting logs.

The answer if you need it for checking is X = 3, Y = 2

Trial and Error? I need STEPS!!! The question is created by me,but can't get the answer by substitution...

Gan it can't be solved by substition because a "base x" and x share only commonality as a number there is no actual mathematical relationship between them


Lets rewrite the equations with two variable substitions

Let A = log 27 base x
Let B = log 20 base y



So equation 1:

3^(log 27 base x) - 2^(log 20 base y) = 7

becomes

3^A - 2^B = 7

equation 2 stays as is

x^y + y^x = 17


See the problem 4 variables 2 equations ... so normally it can't be solved.

However it's your lucky day because equation 1 is a power and we know the numbers 3^A and 2^B a positiver number to any power can only be positive. That restriction makes it solvable because you can work out A & B and therefore you can work out base x and basse y.

You notice I never talked about equation 2 in all the discussion above and for good reason it is pointless because it doesn't help, you only need the top equation to solve the problem.

Its only coincedental that base x and actual x and base y and actual y match.

If you changed the second equatiuon to whatever you like base x and actual x may actually differ because there is no real relationship between them mathematically. The first equation is self consistant and solvable without the need for the second equation.

Science (physics and chemistry in particular) tends to be depth first - the classes deal with very few topics, and beat them to death. There's also a context; everything means something. Science is interesting even when you don't quite know what's going on, because there's some hint of what you're supposed to be doing.

Math is breadth first - it covers a huge number of different things, in a very shallow way. There's no context, it's just a bunch of rules. Because of this ass-backwards method of teaching, it takes a looong time to get to a point where math makes as much sense as science does.