104
Chapter 1 | Functions and Graphs
Rewriting this equation as
e 2 x −5 e x +6=0,
we can then rewrite it as a quadratic equation in e x :
( e x ) 2 −5( e x )+6=0.
Now we can solve the quadratic equation. Factoring this equation, we obtain ( e x −3)( e x −2) =0. Therefore, the solutions satisfy e x =3 and e x =2. Taking the natural logarithm of both sides gives us the solutions x = ln3, ln2.
Solve e 2 x /(3+ e 2 x ) =1/2.
1.30
Example 1.37 Solving Equations Involving Logarithmic Functions
Solve each of the following equations for x . a. ln ⎛ ⎝ 1 x ⎞ ⎠ =4 b. log 10 x +log 10 x =2 c. ln(2 x )−3ln ⎛ ⎝ x 2 ⎞ ⎠ =0
Solution a. By the definition of the natural logarithm function, ln ⎛ ⎝ 1 x ⎞
⎠ = 4if and only if e 4 = 1 x .
Therefore, the solution is x =1/ e 4 . b. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as log 10 x +log 10 x = log 10 x x = log 10 x 3/2 = 3 2 log 10 x .
Therefore, the equation can be rewritten as 3 2 log 10
x =2or log 10 x = 4 3 .
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