Using strong induction (so, neither weak nor "weak++") on the number of matches in the pile, prove that if this number is of the form 4k + 1, for k ∈ N, then P2 always has a strategy to win (which they will follow, because they are smart). Otherwise, P1 always have a strategy to win (which they will also follow because they’re smart too)! To be clear, you will need to prove both facts with a single inductive proof! That is to say, you will need to prove that P2 can always win under the aforementioned condition, but whenever that condition is not met, you have to prove that it is now P1 that can always win!

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like:

if y=4x then you will substitute 4x in for y in the second equation.

8x + 4x = 18 combine like terms

12x = 18 divide by 12 on both sides

x = 18/12 or 3/2

now replace x in the first equation with 3/2 to get

y = 4 3/2

y = 6

the answer is a

(3/2,6)

hope that .

brainliest is always apprectiated