Can you prove 3^x is equal to 3^(1-x) using only expoents rules?
You are not allowed to turn it into an equation
>>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
>>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
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