Can someone explain to me how to do this? : 8^2y+4 = 16^y+1

Answer:y = -4Step-by-step explanation:Assuming the problem is [tex]8^{2y+4}=16^{y+1}[/tex], 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 -- [tex]2^3[/tex] and [tex]2^4[/tex], we can rewrite the original equation by substituting those in:[tex](2^3)^{2y+4}=(2^4)^{y+1}[/tex]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:[tex]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}[/tex]With both of our bases the same, we can now simply compare their exponents directly to solve for y:[tex]6y+12=4y+4\\2y=-8\\y=-4[/tex]