Which of the following is the solution to the equation 8^y + 2) = 2^ y?
a) y = 1 b) y= 3 c) y= -3 d) y= -1
What you can do is plugin each of the possible answers... But I will show you how to solve.
First, you need to distribute the 8: 8(y + 2) = 8y + 16 Ok then lets try your answer and see what we get.
ok then we are stuck with the same problem. Why dont we try inserting the variable and then add before we distribute.
8(y + 2) What do you want to try first?
distribute?
you start with 8(y + 2) = 2y britanny has the right idea: distribute the 8 8y + 16 = 2y now add -2y to both sides 8y -2y + 16 = 2y + -2y simplify 6y + 16 = 0 add -16 to both sides 6y = 16 divide both sides by 6 y = 16/6 = 8/3 are you sure you typed in the correct question ?
now look i added ^ those thangs sorry.
yes, that makes more sense \[ 8^{y+2} = 2^y \] the trick is to know that 8= 2*2*2 , or 2^3 \[ \left(2^3\right)^{y+2} = 2^y \] use the rule that \[ \left(x^a\right)^b= x^{ab} \] to rewrite the problem as \[ 2^{3(y+2)} = 2^y \] now set the exponents equal to each other 3(y+2) = y can you solve this ?
no
3(y+2) = y distribute the 3 on the left side.
3y+6=y ?
yes, now add -y to both sides: 3y+ -y + 6=y + -y can you simplify that ?
3+6=y ? idk that one confused me
3y+ -y + 6=y + -y on the right side you have y + -y which says " 1y take away 1 y" (you get zero) on the left side you have 3 y and take away 1 y can you do it now ?
3y+6=y ?
that is what you start with. now add -y to both sides 3y -y + 6 = y -y what is y-y ?
y-y is nothing? just 0
yes, so 3y -y + 6 = y -y simplifies (on the right side) to 3y - y + 6 = 0 now what is 3 y's take away 1 y ?
2y's
so now you have 2y + 6 =0 next, add -6 to both sides. can you do that ?
2y + 6 -6 = 0-6 or 2y = -6 now divide both sides by 2
-3 is the answer??
yes
thanks
Join our real-time social learning platform and learn together with your friends!