How do you solve #11^ { x - 1} = 1331#?

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Answer 1 · 2018-03-30T19:16:40

# x=4#.

#11^(x-1)=1331#.

#:. 11^x*11^-1=1331#.

#:. 11^x=1331*11=11^3*11^1=11^4#.

#:. x=4#.

This root also satisfy the given equation.

# :. x=4# is the solution.

Answer 2 · 2018-03-30T19:20:28

#color(magenta)(x=4#

#11^(x−1)=1331#

#=> 11^(x−1)=11^3#

Since the bases are same on either sides, we equate the powers.

#=> (x−1)=3#

#=>color(magenta)(x=4#

Answer 3 · 2018-03-30T19:22:07

#11^(x-1)=1331#

#rArrx=4#

We can use the properties of logarithms to solve for #x#.

#11^(x-1)=1331#

#rArrln(11^(x-1))=ln1331#

#rArr(x-1)ln11=ln1331#

#rArrx-1=ln1331/ln11#

#rArrx=ln1331/ln11+1#

This is a legitimate solution, but it can be greatly simplified by recognizing that #1331 = 11^3#

Then we can rewrite the equation as:

#rArrx=ln(11^3)/ln11+1#

#rArrx=(3ln11)/ln11+1#

#rArrx=3+1#

#rArrx=4#

It now seems obvious that if we plug #x=4# back into the original expression...

#11^(x-1)=11^(4-1)=11^3=1331#