logarithms

Question

solve and explain

 

(1/6)^x-1 + 4 = 40

Answer 1

(1/6)^x-1 +4 = 40

(1/6)^x-1 = 36   (subtracting 4 from both sides)

(6^-1)^x-1  = 6² (  1/6 = 6^-1 , 6² =6 × 6 =36)

6^-1(x-1)  =  6²

-1(x-1) = 2  (If a^x = a^y , then x = y)

-x+1 = 2 (distributive property)

-x = 1 (subtracting 1 from both sides)

x = -1 (multiplying both sides by -1)

Answer 2

    (1/6)^x-1 + 4 = 40

                 - 4     -4

 

 

    (1/6)^x-1 = 36

 

Take the log (any base) of each side of the equation - I'll use natural log:

 

    (x-1)*ln(1/6) = ln(36)

 

    x-1 = ln(36)/ln(1/6)

 

    x = ln(36)/ln(1/6) + 1

 

    x = -1

 

Answer 3

(1/6)^x-1 = (6^-1)^x-1 = 6^1-x

 

So, your equation is equivalent to:  6^1-x + 4 = 40

                                                         6^1-x = 36

                                                         6^1-x = 6²

                                                           1-x = 2

                                                              x = -1