Simplify giving the answer in exponential form.(6)^5^7 =(6)^p-3

(6)^5^7=(6)^p-3

Here  on the right side of the equation, -3 is shown without being enclosed in bracket.The question has an unknown P. To simplify the left side it is a very large number. So I take cases where  only p is the power of 6 on right side and -3 is not and then taking (p-3) together in exponent.

There is no simplification. It is an equation involving  an unknown p and the question should therefore be  to solve for P.

I do not understand the purpose of asking the answer in the exponential form by a 9th grade student. But I give the solution for p in the exponential form  (of course no base you mentioned.)

Case 1:

By, the order of priority, (BODMAS),the leftside  has 2 exponents of equal priority. So the operation of exponents in (6)^5^7  is 6^5 and the another power 7 and so, (6)^5^7=6^35.

(6)^35 = (6)^p-3 =>

(6)^35+3=6^p

Taking logarithms,

P log 6 = log(6^5^7+3)

P=log(6^5^7+3)/ log6=(1/log6) log (6^35)[1+ 1/6^35]

=35( log6/log6)+ log(1+1/6^35).

=35+ 0 as log(1+1/large quantity)->log 1 =0

Therfore, P =35  =  6^1.984277534 in exponential form

Case 2:

However, if you think 5^7 has to be decided first , then

6^p=6^(5^7)+3

P=log [6^(5^7)+3]/log6 = 5^7 +log[1+1/6^(5^7)]

=5^7+0

=5^7 in exponential form of base 5 .

Case 3a:

If you intend (6)^5^7 =6^(p-3), then bases being equal, powers should be equal.

Therefore 6^35= 6^(p-3), Equating powers,

P =35+3=38 = 6^2.030175491,in exponential form.

Case 3b:

(6)^(5^7) = 6^(p-3), then p-3=5^7 or p=5^7+3