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

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

I believe the equation you want to be solved is:

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

And I believe your question should correctly read "simplify and give answer." Giving answer to or solving an equation usually refers to finding the value of the unknown variable, which in this case is 'p'. Whatever is the value of 'p' is the answer. It does not mean much to say that give value in exponential form, because multiple answer are possible depending on the base number used.

Now we will get down to solving the problem itself. To simplify the left hand side of the equation we will use the following rule:

A^B^C = A^(B*C)

Therefore: (6)^5^7 = (6)^(5*7) = (6)^35

Therefore: as per the equation given  (6)^35 = (6)^(p-3)

Therefore: 35 = (p-3)

Therefore: p = 35+3 = 38  ... Answer

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

(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

 

 

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