How do you solve (7^-1)^-1 and 5^2 * 5^4 * 5^-3

Step-by-step explanation:

exponents raised to a - power  is an inverse function probably not the right word

       a^-b    =  1/(a^b)     example       4^-2   = 1/(4^2)     =  1/16

so   (7^-1)^-1  is a double inverse which is back to the original number  7

      inverted 7 twice

      (7^-1)^-1     = (1/7)^-1     =  1/(1/7)     =   7     trust me or use your calculator

since the coeffients are the same  just add the powers...  some exponent rule

     5^2 * 5^4 * 5^-3   = 5^(2+4-3)   = 5³

here is it written out

5^2 * 5^4 * 5^-3    = 5²  ×   5^4   ×  5^-3    =   (5×5) × (5×5×5×5)  × (1/5×1/5×1/5)

    5×5 × 5× (5×5×5  × 1/5×1/5×1/5)           (5×5×5  × 1/5×1/5×1/5)  = 1

    5×5×5    = 5³

   so

you are saying it as if a fact, not a question, but yes, you would be correct? ; )

step-by-step explanation: