Question

Math help

Answer 1

@Zepdrix

Answer 2

pardon Zepdrix but eviant this exercise i ve solved with you yestoday - than i remember right

Answer 3

@jhonyy9 I don´t understand how to find the base on my own

Answer 4

That last option is kinda strange...
I don't see why having a negative coefficient in front would make it NOT an exponential. That's a weird option. Maybe they were referring to a negative base value.

But ignoring all that,
\[\large\rm -\left(\frac43\right)^{-6x}\quad=\quad-\left[\left(\frac43\right)^{-6}\right]^x\quad=\quad -\left[\color{royalblue}{\left(\frac34\right)^6}\right]^x\]

Ya that first option looks about right :D

Oh you guys already do this one?

Answer 5

I want to be able to solve my math problems on my own

Answer 6

I first applied the "power rule" for exponentials,\[\large\rm x^{ab}=(x^a)^b\]which allowed me to pull the -6 and x apart.

Then applying our "flip" exponential rule, or whatever you like to call it,\[\large\rm x^{-a}=\frac{1}{x^a}\]

Answer 7

@Zepdrix how did the -6 become a positive?

Answer 8

Zepdrix my opinion that this negativ exponent make the base being this fraction inversed
hope you understand me

Answer 9

It's one of your exponent rules.
\[\large\rm \left(\frac{x}{y}\right)^{-a}\quad=\quad \left(\frac{y}{x}\right)^a\]Negative exponent goes away, and fraction becomes the reciprocal, it flips.

Answer 10

oh, thankx

Answer 11

very nice congrat Zepdrix for explication - ty.