Pls halp i am autismo i should never have taken maths but my parents will probably kill themselves if i become anything...

pls halp i am autismo i should never have taken maths but my parents will probably kill themselves if i become anything other than an engineer so if not for me help me for their sake

3/2

-2/3

most likely -(2/3)

this guy (math teacher here)
Negative two thirds

This

log(all the shit on the left side) / log(5)

I'm blanking on the operation...when does the fraction become negative?

19407837821

fucking unhelpful answers.

n = log5(left side of equation)

You should be able to solve this.

Call this number

>Multiply both sides by the denominator, 5^2/3
>Multiplying like bases you add the powers so we get 1 = 5^n+(2/3)
>log both sides to get 0=n+ (2/3)
>subtract 2/3
>n=- 2/3

an denominator can be written as being a numerator to a negative power even if that power is -1

1/2 = 2^(-1)

The bottom part of the fraction is equal to 5^(2/3) because a root is the inverse of a index.

The entire fraction is 1/(5^(2/3)), which is 5^(-2/3) due to reciprocal laws

So you now have 5^(-2/3)=5^n, so it should be obvious now that n=-2/3

1/x = x^(-1)
third_sqrt(x) = x^(1/3)

--

1/(third_sqrt(5))^2 = third_sqrt(5)^-2 = (5^(1/3))^-2 = 5^(-2/3)

=> n = -2/3

...

You stupid fuck.

The cube route of 5 is 5^(1/3)
Square that 5^(2/3)
Reciprocal 5^(-2 / 3)
n = - 2 / 3
Fucking newfag. Post nudes.

-2/3

this is the simplest explanation so far

>Implying being a newfag is somohow linked to not being able to solve an equation

1/((cube root(5))^2) = 1/((5^(1/3))^2) = 1/(5^(2/3)) = 5^(-2/3)

>n = -2/3