Indices

Because $49 = 7^2$

$49 \cdot 49 \cdot 49 \cdot 49 = 49^4 = (7 ^2)^4$

Since $(n^x)^y = n^{xy}$

$(7^2)^4 = 7^8 = 7^a$

$a = 8$

49 x 49 x 49 x 49=7^2 x 7^2 x 7^2 x 7^2

=>7^8=7^a

since bases are same, we can compare the powers

=>a=8

$\begin{aligned} 49 × 49 × 49 × 49 & = 7^2 × 7^2 × 7^2 × 7^2\\ & = 7^{2+2+2+2}\\ & = 7^8\\ \implies a & = \color{#69047E}{\boxed{8}} \end{aligned}$

Square root of 49 is 7 (7*7=49) there is the number 49 8 times, hence it makes an exponent sum and the answer coclutes that is 8

7^a=49x49x49x49=7^2x7^2x7^2x7^2=7^(2+2+2+2)=7^8. or, a=8

that was quite easy as we know that7^2=49.

49x49x49x49=7a,7(7)x7(7)x7(7)x7(7)=7^8 a=8

49 can be written as 7square. so we have four 49s...and we get (7)*8 so a=8

We have

$49 = 7^2 \implies 49^4=(7^2)^4=7^8,$

which yields $a=\boxed{8}$ .