$40^8 \cdot 80^4 = 2^x \cdot 5^y$ -Finding the $x+y$

Question

$$40^8 \cdot 80^4 = 2^x \cdot 5^y$$

• Find the $x+y$

So, this question seems really too easy. However, the thing is to know where to start. Let me explain what I thought. I see that $2$ and $5$ are both divisor of $40$ and $80$. Also, personally I'd like to ask an anohter question as well.

• Is there any strategy to solve these equalities which involve exponential terms?

Regards!

Answer 1

$ 2^x \cdot 5^y=40^8 \cdot 80^4 =(2^35)^8(2^45)^4=2^{40}5^{12}$

$x=40$ and $y=12$.

Answer 2

$$40^8 \cdot 80^4 = 2^x \cdot 5^y\\ { \left( { 2 }^{ 3 }\cdot 5 \right) }^{ 8 }\cdot { \left( { 2 }^{ 4 }\cdot 5 \right) }^{ 4 }={ 2 }^{ x }\cdot { 5 }^{ y }\\ { 2 }^{ 24+16 }\cdot { 5 }^{ 8+4 }={ 2 }^{ x }\cdot { 5 }^{ y }\\ x=40,y=12 $$

Answer 3

$(2\cdot 2\cdot 2\cdot 5)^8 \cdot (2\cdot 2\cdot 2\cdot 2\cdot 5)^4 = 2^{40} \cdot 5^{12}$