If 3^p+3^4=90, 2^r+44=76, and 5^3+6^s=1421, what is the product of p, r, and s?

2 Answers
Mar 25, 2018

Given 3^p+3^4=90, 2^r+44=76, and 5^3+6^s=1421,

So 3^p+3^4=90

=>3^p=90-81=3^2

=>p=2

Again

2^r+44=76

=>2^r=76-44=32=2^5

=>r=5

Again

5^3+6^s=1421

=>6^s=1421-125=1296=6^4

=>s=4

Hence pxxrxxs=2xx5xx4=40

Mar 25, 2018

prs=40

Explanation:

As 3^p+3^4=90, we have 3^p=90-3^4=90-81=9

i.e. 3^p=3^2 and hence p=2

Similarly 2^r+44=76 implies 2^r=76-44=32=2^5 and

hence r=5

and 5^3+6^s=1421 implies 6^s=1421-5^3=1421-125=1296=6^4

and therefore s=4

hence prs=2*5*4=40