Solving these equations requires using algebraic manipulation and the properties of exponential and logarithmic functions to isolate the variable in the equation.
Solve: 24x = 22x + 1
4x = 2x + 1 using the identity property 2x = 1 x = 1/2
Solve: 22x = 8x + 1
22x = (23)x+1 = 23x + 3 change to a common base 2x = 3x + 3 x = -3
Solve by taking logarithm of both sides: 2x = 16
lg 2x = lg 16 = 4 use the base of the exponential x = 4
Solve by taking logarithm of both sides: ex = 8
ln ex = ln 8 use natural logarithm base x = ln 8 = 2.079442
Solve by taking logarithm of both sides: 102x = 200
log 102x = log 200 use base 10 log 102x = 2x log 200 = log 2 + log 100 = log 2 + 2 2x = log 2 + 2 x = 0.5 log 2 + 1 = 1.150515
Solve by taking logarithm of both sides: 3x + 1 = 5x
log 3x + 1 = log 5x arbitrary base
(x + 1) log 3 = x log 5
log 3 = x log 5 - x log 3 = x(log 5 - log 3)
log 3 0.477121
x = ------------- = ------------------- = 2.150660
log 5 - log 3 0.698970 - 0.477121
It is possible that a potential solution obtained by isolating the variable results in taking a logarithm of a number outside the logarithm domain (0, +∞). If this happens, there is no solution to the equation. For example, try to solve:
log 2x = log (x - 1) 2x = x - 1 using the identity property x = -1 By substituting the result into the original equation, we get: log (-2) = log (-2) Domain error!
Solve: log (x + 2) + log (x - 2) = log 3x
log [(x + 2)(x - 2)] = log 3x x2 - 4 = 3x x2 - 3x - 4 = 0 (x - 4)(x + 1) = 0 Potential solutions: x = -1 or 4 Substitute x = -1 into equation: log (-1 + 2) + log (-1 - 2) = log (-3) Domain error! Substitute x = 4 into equation: log (4 + 2) + log (4 - 2) = log 12 log 6 + log 2 = log 12 Ok! Solution is: x = 4
Solve: ln 4 - ln (5 - x) = ln x
ln [4/(5 - x)] = ln x 4/(5 - x) = x 4 = x(5 - x) 4 = 5x - x2 x2 - 5x + 4 = 0 (x - 1)(x - 4) = 0 x = 1 or 4 Substitute x = 1: ln 4 - ln (5 - 1) = ln 1 ln 4 - ln 4 = ln 1 Ok! Substitute x = 4: ln 4 - ln (5 - 4) = ln 4 ln 4 - ln 1 = ln 4 ln 4 - 0 = ln 4 Ok! Solution set: x = 1, 4
Some logarithmic equations can be solved by converting to an equivalent exponential form. For example, solve:
2 = log (3x + 1) 102 = 3x + 1 using base 10 3x = 99 x = 33 Check result: log (3 × 33 + 1) = log 100 = 2 Ok!
Solve: lg (x + 4) + lg (x + 12) = 7
lg [(x + 4)(x + 12)] = 7 x2 + 16x + 48 = 27 = 128 x2 + 16x - 80 = 0 (x - 4)(x + 20) = 0 Check: x = 4 lg (4 + 4) + lg (4 + 12) = lg 8 + lg 16 = 3 + 4 = 7 Ok! Check: x = -20 lg (-20 + 4) + lg (-20 + 12) Domain error! Solution: x = 4
Solve: ln x = 1 + log x
ln x = log x / log e change base of left hand side
log x = (log e)(1 + log x)
log x - (log e)(log x) = log e
(1 - log e)log x = log e
log e 0.434294
log x = --------- = -------------- = 0.767704
1 - log e 1 - 0.434294
x = 100.767704 = 5.857390
Check result:
ln 5.857390 = 1 + log 5.857390
1.767704 = 1 + 0.767704 Ok!
Solution: x = 5.857390