Solving Exponential Equations: 2 Problems with Multiple Solutions

[EternityMech]

Homework Statement

it wants the sum of all solutions

2 different problems 1. and 2.

1. 2^x + 2^(x+1) + 2^(x+2) = 2^6.

2. 8/(3^x +2)>=3^x

>= meaning equal or greater than 3^x

Homework Equations

should i come up with a summa equation?

The Attempt at a Solution

i switched 3^x=y and solved it but i only got 1 solution and apparently there are more. if anyone can help. thanks.

 

[eumyang]

1. 2^x + 2^(x+1) + 2^(x+2) = 2^6.

Note that
2^{x+2} = 2^2 \cdot 2^x = 4 \cdot 2^x
and
2^{x+1} = 2^1 \cdot 2^x = 2 \cdot 2^x.

So
2^x + 2^{x+1} + 2^{x+2} = 2^x + 2 \cdot 2^x + 4 \cdot 2^x = 7 \cdot 2^x.
Take it from there.

 

[Mark44]

Homework Statement

it wants the sum of all solutions

2 different problems 1. and 2.

1. 2^x + 2^(x+1) + 2^(x+2) = 2^6.

2. 8/(3^x +2)>=3^x

>= meaning equal or greater than 3^x

Homework Equations

should i come up with a summa equation?

The Attempt at a Solution

i switched 3^x=y and solved it but i only got 1 solution and apparently there are more. if anyone can help. thanks.

In #1, the equation is the same as 2^x + 2*2^x + 4*2^x = 2⁶. Can you solve that one?

In #2, 3^x > 0 for all x, so 3^x + 2 > 2 for all x.
Multiplying both sides of the inequality by 3^x + 2 won't change the direction of the inequality. If you do this, you get an inequality that is quadratic in form. Can you show us what you did?