>>7790782
>can you prove 3^x is equal to 3^(1-x) using only expoents rules?

Let x=0
3^0=1
3^(1-0)=3^1=3

1=/=3

This doesn't even make sense

3^2=9, 3^(1-2)=1/3

>>7790782

no...

x=1/2

This is possible the solution is 1/2. But I need to know how to solve it using only expoents rules. Dont make an equation out of it, making 3^x= 3^(1-x) is the easy way

You cant just give out values to x. X only has one solution and that already was found out so dont say it doesnt make sense just because the solution is not what you want

rule 2
rule 1 + 5
2x = 1

fill in gaps as necessary

>>7790853
Checking it

>>7790782
>prove 3^x is equal to 3^(1-x)
>prove
no one can because it is wrong

>>7790853
Doesnt make sense

3^x = 0^x : 3^x wont get you to 3^(1-x)

Nor the other way around : 3^(1-x) = 3^1 : 3^x as this isnt nowhere near 3^x

None of the above work

>>7790863
Its not wrong when x=1/2 therefore solution is 1/2. I dont think you can get to x using only the expoents rules

>>7790853
>>7790869

rule 2: 3^(1-x)=3^1/3^x

(multiplying 3^x and 3^1/3^x by 3^x)

rule 1: 3^x*3^x = 3^(x+x) = 3^(2x)
rule 5: 3^1*(3^x/3^x) = 3^1*(3/3)^x = 3^1

2x = 1

>>7790874
>>7790782
yet you said that you want a "proof", not a "solution". this particular problem can be solved, but cannot be proven because, as someone already said, it is not a true statement.

>>7790896
I might have not been fully explicit. The exercise says that 3^(1-x)=6 and you have to take 3^x turn it into 3^(1-x) so you can replace it by 6 and calculate whatever expression you have so you can get 1/2

You should only do it using the rules and thats what Im having problems with

>>7790920
so do you mean like 3^(1-x) = 3^1 * 3^(-x) = 3^1 * 1/3^x
now you have an expression containing 3^x so you can substitute.

that's how I would interpret the instructions.

>>7791023
You interpreted it right the only problem is that you took the wrong expression. Do the same starting with 3^x

Take this example:

You know that 3^(1-x) = 6. The expression in the example is 3^(2-2x)

Solution:

3^(2-2x)= 3^((1-x)*2) [rule 3] = 6^2 = 36

>>7791023
3^x = 3^(x-1+1) = 3^(x-1) * 3 = 1/3^(1-x) * 3

>>7791155 was meant as a reply for >>7791133

>>7791155
Genius. Thanks, its exactly this

>>7790874
What they mean is that it has been worded incorrectly.
A proof that 3^x is equal to 3^(1-x) is a proof that they are equal for ALL x. What should have been asked for is a solution to 3^x = 3^(1-x), or in words:
>Can you find the values of x for which 3^x is equal to 3^(1-x), using only these exponent rules?

>>7791155
Where did that 1/3 come from?

>>7791225

You're right. My mistake