6^{x+2}\cdot 3^x\div 2^{5-x}=3^5

Answer 1

6^x+2 · 3^x ÷ 2^5-x = 3⁵

We have three bases in our equation: 6, 3, and 2. We can utilize the fact that 6 = 3 · 2 to rewrite 6^x+2 as (3 · 2)^x+2, which expands to 3^x+2 · 2^x+2. The left side of our equation now reads 3^x+2 · 2^x+2 · 3^x ÷ 2^5-x . Since multiplication is commutative, we can swap the two middle terms to get the expression 3^x+2 · 3^x · 2^x+2 ÷ 2^5-x. We can use the following properties of exponents to help simplify this:

a^x · a^y = a^x+y

a^x ÷ a^y = a^x-y

Applying those properties gives us the simplified expression 3^x+2+x · 2^x+2-5+x = 3^2x+2 · 2^2x-3, which we can take to our original equation to get one far neater-looking than the one we started with: 3^2x+2 · 2^2x-3 = 3⁵. Diving both sides of the equation by 3⁵ gives us a 1 on the right side and, using the division property of exponents, makes the left side 3^2x+2-5 · 2^2x-3 = 3^2x-3 · 2^2x-3. For 3^2x-3 · 2^2x-3 to equal 1, both terms in the product must also equal one: their powers must be 0. This is equivalent to saying 2x - 3 must equal 0, which we can solve fairly easily for x: 2x - 3 = 0 → 2x = 3 → x = 2/3.

Our answer: x = 2/3