Find the value of x if:
y × y × y ÷ y³ × y^4 = y²^x
pls tell step by step
Answers
Answer:
y^4 = y*y*y*y
y^3 = y*y*y
y^2 = y*y
Now, when computing y^3 * y^3 we can just do:
y^3 * y^3 = (y*y*y) * (y*y*y) = y*y*y*y*y*y = y^6
Next, when computing y^2 * y^4 we can just do:
y^2 * y^4 = (y*y) * (y*y*y*y) = y*y*y*y*y*y = y^6
Finally, we have the equation:
y^2 * y^4 =y^6
y^3 * y^3 + y^2 * y^4 = y^6 + y^6 = y^12
y^6/y^6=1
the 2 y’s cancel out as well as the power symbol leaving us with 6/6=1.
when the 1’s cancels out as well as the power symbol you are left with y/y=y which will be equivalent to 6/6 which will equal 1 making y equal 1.
Y^2=y*y
y^6=y*y*y*y*y*y
y^8=y*y*y*y*y*y*y*y
y*y+ y*y*y*y*y*y+y=y*y*y*y*y*y*y*y*y(y^8+y)
this is y^2+y^6+y which equals y^8+y which in our case is y^9 but in other cases it wouldn’t be the same since it would have a higher value at the end when you add y than if you timesed by all the values of y and in other cases when you times by all the values of y it would be bigger than just adding y on, after you times y.
Y^2=y*y
y^6=y*y*y*y*y*y
the Simplified numbers are 1 i.e y^1
the 2nd solution is to add ^1 to all y^3, and add ^2 to y^2 so they all are consistent and can be simplified.
y^3+^1=y^4
y^3+^1=y^4
y^4*y^4+y^4*y^4/4=y^1*y1+y^1*y^1