Question 1036472
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Solve for x 3^x+1 - 4 + 1/3^x =0
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I will read it as

{{{3^(x+1) - 4}}} + {{{1/3^x}}} = {{{0}}}.

Rewrite it in this way

{{{3*3^x - 4}}} + {{{1/3^x}}} = {{{0}}}.    (1)

Introduce new variable u = {{{3^x}}}. Then the equation (1) takes the form

{{{3u -4 + 1/u}}} = {{{0}}}.

Multiply both sides by "u" to rid of denominator. You will get

{{{3u^2 - 4u + 1}}} = {{{0}}}.

Factor the left side:

(3u-1)*(u-1) = 0.    (2)

Then the equation (2) splits in two independent equations:


1)  3u - 1 = 0  --->  u = {{{1/3}}}.  Then  {{{3^x}}} = {{{1/3}}}.  Hence, x = -1 is the solution of the original equation (1).

2)  u - 1 = 0  --->  u = 1.  Then {{{3^x}}} = 1.  Hence, x = 0 is the solution of the original equation (1).

<U>Answer</U>.  The solutions of the original equation are x = 0  and  x = -1.
</pre>

The lesson to learn from this solution is the method of introducing a new variable.