Question

Can someone explain to me how to do this? :

8^2y+4 = 16^y+1

Answer 1

y = -4

Explanation:

Assuming the problem is
`8^(2y+4)=16^(y+1)`, it would be nice if we could convert both sides of this equation to the same base; that way, we could compare the exponents directly in an equation of their own. Fortunately, 8 and 16 are both powers of 2 --
`2^3` and
`2^4`, we can rewrite the original equation by substituting those in:

`(2^3)^(2y+4)=(2^4)^(y+1)`

When you have an exponent raised to another exponent, you multiply those exponents together, so we can simplify our equation by distributing a 3 in the left exponent and a 4 in the right:

`2^(3\cdot(2y+4))=2^(4\cdot(y+1))\\2^(3\cdot2y+3\cdot4)=2^(4\cdot y+4\cdot1)\\2^(6y+12)=2^(4y+4)`

With both of our bases the same, we can now simply compare their exponents directly to solve for y:

`6y+12=4y+4\\2y=-8\\y=-4`