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==Solving Logarithmic Equations== To solve logarithmic equations we take advantage of the properties of logarithmic functions and the fac..."

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==Solving Logarithmic Equations==

To solve logarithmic equations we take advantage of the properties of logarithmic functions and the fact that

<math> y = log_a(x)\text{ is equivalent to } x = a^y ~ a > 0,~ a \neq 1 </math>

We also use the additional fact that if <math>log_a(M) = log_a(N) </math> then M = N for M, a, N positive numbers and <math> a \neq 1</math>

==Example==
Solve: <math> log_5(x+ 6) + log_5(x + 2) = 1</math>

::<math>\begin{array}{rcl}
log_5(x + 6) + log_5(x + 2) & = & 1\\
log_5( (x + 6)(x + 2)) & = & 1\\
(x + 6)(x + 2) & = & 5\\
x^2 + 8x + 12 & = & 5 \\
x^2 + 8x + 7 & = & 0\\
(x + 1)(x + 7) & = & 0
\end{array}</math>

Now we just need to make sure our answers make sense. When x = -7, we have <math>log_5(-1) + log_5(-5)</math> which cannot occur since the domain of
the logarithm function is <math>(0, \infty)</math>

==Solving Exponential Equations==

In a similar fashion to solving logarithmic equations, we can solve exponential equations by using their properties and the fact that if <math> a^u = a^v ~\text{ then } u = v~ a > 0, ~a \neq 1</math>

Example:

Solve: <math> 8\cdot 3^x = 5</math>

We start by dividing both sides by 8 to get <math> 3^x = \frac{5}{8}</math>. Taking the log base 3 of both sides we find that <math> log_3(3^x) = log_3(\frac{5}{8})</math>.
Finally by our properties of logarithms <math>x = log_3(\frac{5}{8})</math>.

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Logarithmic and Exponential Equations - Revision history

MathAdmin: Created page with "__TOC__ ==Solving Logarithmic Equations== To solve logarithmic equations we take advantage of the properties of logarithmic functions and the fac..."

MathAdmin: Created page with "
TOC
==Solving Logarithmic Equations== To solve logarithmic equations we take advantage of the properties of logarithmic functions and the fac..."