exp_log_equations

exp_log_equations - Equations involving exponential and...

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Equations involving exponential and logarithm functions Example 1 : Solve for x : = 8 ( ) - 3 x 4 128. Solution : = 8 ( ) - 3 x 4 128 Both sides of the equation can be expressed as a power of 2 to give: = ( ) 2 3 ( ) - 3 x 4 2 7 , that is, = 2 ( ) - 9 x 12 2 7 . Since both sides are a power of 2, we can equate the powers. This amounts to taking logarithms to base 2 of both sides of the equation. = - 9 x 12 7. Hence = 9 x 19, so that = x 19 9 . _____ Example 2 : Solve for x : = 3 ( ) 5 x 9 ( ) - 2 x 1 . Solution : = 3 ( ) 5 x 9 ( ) - 2 x 1 Both sides of the equation can be expressed as a power of 3 to give: = 3 ( ) 5 x ( ) 3 2 ( ) - 2 x 1 , that is, = 3 ( ) 5 x 3 ( ) - 4 x 2 . Since both sides are a power of 3, we can equate the powers. This amounts to taking logarithms to base 3 of both sides of the equation. = 5 x - 4 x 2. Hence = x - 2. _____ Example 3 : Solve for x : = 16 ( ) x 2 8 ( ) - 5 x 3 . Solution : = 16 ( ) x 2 8 ( ) - 5 x 3 Both sides of the equation can be expressed as a power of 2 to give: = ( ) 2 4 ( ) x 2 ( ) 2 3 ( ) - 5 x 3 , that is, = 2 ( ) 4 x 2 2 ( ) - 15 x 9 . Since both sides are a power of 2, we can equate the powers. This amounts to taking logarithms to base 2 of both sides of the equation. = 4 x 2 - 15 x 9. Collecting all the terms on the left-hand side of the equation gives: = - + 4 x 2 15 x 9 0 ------- (i). Using the quadratic formula gives: = x 15 +/- - 225 144 8 = 15 +/- 81 8 = 15 +/- 9 8 = 15 8 + 9 8 .
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Hence = x 3 or = x 3 4 . Alternatively, the left-hand side of equation (i) can be factored to give: = ( ) - 4 x 3 ( ) - x 3 0, so that = x 3 or = x 3 4 . _______ Example 4 : Solve for x : = 5 ( ) 2 x 16. Solution : = 5 ( ) 2 x 16 Since the left and right-hand sides cannot easily be converted to powers of the same base, we take logarithms of both sides. The natural logarithm function (logarithm to base e ) or common logarithms (logarthms to base 10 ) are both suitable for this. = ( ) log 10 5 ( ) 2 x log 10 16. Making use of the power property of logarithms gives: = 2 x ( ) log 10 5 log 10 16. Hence = x log 10 16 2 log 10 5 ~ 0.861353. ________ Approximate check. When = x 0 .861353, = 5 ( ) 2 x 5 1.722706 ~ 16.0000. Example 5 : Solve for x : = 2 ( ) + x 2 3 ( ) + 2 x 1 . Solution : = 2 ( ) + x 2 3 ( ) + 2 x 1 Since the left and right-hand sides cannot easily be converted to powers of the same base, we take logarithms of both sides. The natural logarithm function (logarithm to base e ) or common logarithms (logarthms to base 10 ) are both suitable for this. = ( ) ln 2 ( ) + x 2 ( ) ln 3 ( ) + 2 x 1 . Making use of the power property of logarithms gives: = ( ) + x 2 ln 2 ( ) + 2 x 1 ln 3. Multiplying out the brackets on the left and right-hand sides gives: = + ( ) ln 2 x 2 ln 2 + ( ) 2 ln 3 x ln 3. Collecting the terms with the factor
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exp_log_equations - Equations involving exponential and...

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