Question

hey frnds can u help in this one:

7^x-1 =1


solve exponents but don't do it by log method..

Answer 1

❤7^x-1=1
7^x=1+1
7^x=2

Answer 2

7^(x-1)= 1

7^ ( x-1) = 7^0

compare

x-1 = 0

x= 1

if it is 7^x -1 = 1

7^x = 1+1 = 2

7^x = e^ln 2

e^ ln ( 7^x) = e^ln2

e^ x ln 7 = e^ln2

compare

x ln 7 = ln2

x= ln 2/ ln 7