how to solve equations with logarithms ♡

to solve an equation that has logarithms, we need to always keep in mind that the domain of the log function only allows values greater than zero
because the equation $a = b$ is the same as $c^{a} = c^{b}$ , we can use this to solve logarithms
we also need to substitute each solution into the equation to check that it is valid

worked example: solve $lo g_{2} (x) + lo g_{2} (x + 6) = 4$
$lo g_{2} (x^{2} + 6 x) = 4∣ x > 0, x + 6 > 0$
$2^{l o g_{2} (x^{2} + 6 x)} = 2^{4} ∣ x > 0, x + 6 > 0$
$x^{2} + 6 x - 16 = 0∣ x > 0, x + 6 > 0$
$(x + 8) (x - 2) = 0∣ x > 0, x + 6 > 0$
$x = - 8 \lor x = 2∣ x > 0, x + 6 > 0$
reject $x = - 8$
$x = 2$

worked example: solve $lo g_{3} (x - 1) + lo g_{3} (x + 1) = 1$
$lo g_{3} (x^{2} - 1) = 1∣ x - 1 > 0, x + 1 > 0$
$3^{l o g_{3} (x^{2} - 1)} = 3^{1} ∣ x - 1 > 0, x + 1 > 0$
$x^{2} - 1 = 3∣ x - 1 > 0, x + 1 > 0$
$x = \pm 2∣ x - 1 > 0, x + 1 > 0$
reject $x = - 2$
$x = 2$

worked example: solve $lo g_{2} (2 x + 1) = lo g_{2} (x - 1) + 4$
$lo g_{2} (2 x + 1) - lo g_{2} (x - 1) = 4$
$lo g_{2} (\frac{2 x + 1}{x - 1}) = 4∣2 x = 1 > 0, x - 1 > 0$
$\frac{2 x + 1}{x - 1} = 2^{4} ∣2 x = 1 > 0, x - 1 > 0$
$\frac{2 x + 1}{x - 1} = 16∣2 x = 1 > 0, x - 1 > 0$
$2 x + 1 = 16 x - 16∣2 x = 1 > 0, x - 1 > 0$
$14 x = 17∣2 x = 1 > 0, x - 1 > 0$
$x = \frac{17}{14}$

worked example: solve $lo g_{10} (20 x) = lo g_{10} (x - 8) + 2$
$lo g_{10} (\frac{20 x}{x - 8}) = 2∣20 x > 0, x - 8 > 0$
$\frac{20 x}{x - 8} = 1 0^{2} ∣20 x > 0, x - 8 > 0$
$20 x = 100 x - 800∣20 x > 0, x - 8 > 0$
$x = 10$

worked example: solve $1 - lo g_{10} (2 x + 1) = lo g_{10} (5 x + 8) - 2 lo g_{10} (x + 1)$
$lo g_{10} (5 x + 8) - 2 lo g_{10} (x^{2} + 2 x + 1) + lo g_{10} (2 x + 1) = 1$
$lo g_{10} (5 x + 8) - lo g_{10} (x^{2} + 2 x + 1) + lo g_{10} (2 x + 1) = 1$
$lo g_{10} (\frac{( 5 x + 8 ) ( 2 x + 1 )}{x ^{2} + 2 x + 1}) = 1$
$\frac{( 5 x + 8 ) ( 2 x + 1 )}{x ^{2} + 2 x + 1} = 10$

worked example: solve $lo g_{2} (x + 3) + lo g_{2} (x - 1) = 4$
$(x + 3) (x - 1) = 16∣ x + 3 > 0, x - 1 > 0$
$x^{2} + 2 x - 3 = 16∣ x + 3 > 0, x - 1 > 0$
$x^{2} + 2 x - 19 = 0∣ x + 3 > 0, x - 1 > 0$
quadratic formula
$x = - 1 \pm 25 ∣ x + 3 > 0, x - 1 > 0$
$x = - 1 + 25$

worked example: solve $2 lo g_{3} (x + 1) = 3 + lo g_{\frac{1}{3}} (1 + x)$
$lo g_{3} (x^{2} + 2 x + 1) = 3 + lo g_{\frac{1}{3}} (x + 1) ∣ x + 1 > 0$
$lo g_{3} (x^{2} + 2 x + 1) = 3 + \frac{lo g _{3} ( x + 1 )}{lo g _{3} ( \frac{1}{3} )} ∣ x + 1 > 0$
$lo g_{3} (x^{2} + 2 x + 1) = 3 + \frac{lo g _{3} ( x + 1 )}{- 1} ∣ x + 1 > 0$
$lo g_{3} (x^{2} + 2 x + 1) = 3 - lo g_{3} (x + 1) ∣ x + 1 > 0$

$lo g_{3} ((x^{2} + 2 x + 1) (x + 1)) = 3∣ x + 1 > 0$
$(x^{2} + 2 x + 1) (x + 1) = 27∣ x + 1 > 0$
$(x + 1)^{3} = 27∣ x + 1 > 0$
$x + 1 = 3∣ x + 1 > 0$
$x = 2$

worked example: consider the function $f (x) = 2^{1 - x}$
simplify $\frac{f ( x )}{f ( - x )}$ into the form $2^{k x}$
$\frac{2 ^{1 - x}}{2 ^{1 + x}}$
$2^{1 - x - 1 - x}$
$2^{- 2 x}$

show that $f (a) \cdot f (b) = f (a + b - 1)$
$2^{1 - a} \cdot 2^{1 - b} = 2^{1 - a - b + 1}$
$2^{1 - a - b + 1} = 2^{1 - a - b + 1}$

solve $f (x) = 3$ for $x$ , give answer in the form $x = lo g_{2} (\frac{m}{n}) + c$
$2^{1 - x} = 3$
$1 - x = lo g_{2} (3)$
$- x = lo g_{2} (3) - 1$
$x = - lo g_{2} (3) + 1$
$x = lo g_{2} (3^{- 1}) + 1$
$x = lo g_{2} (\frac{1}{3}) + 1$

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how to solve equations with logarithms

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