03Dlogarithmiceqns

toc = Logarithmic equations =

We can solve logarithmic equations by rewriting them in index form.

**Example 1**
Solve for **//x//**: math \\ (a) \quad \log_2{(2x-1)}=4 \\ \\ (b) \quad \log_3{(3x+1)}=0 \\ \\ (c) \quad \log_3{\big(\frac{1}{9}\big)}=x math

__Solution:__

math \log_2{(2x-1)}=4 \;\; \iff \;\; 2^4=2x-1 math
 * (a) **

math \\ 2x-1=16 \\ \\ 2x=17 \\ \\ x=\dfrac{17}{2} math

math \log_3{(3x+1)}=0 \;\; \iff \;\; 3^0=3x+1 math
 * (b) **

math \\ 3x+1=1 \\ \\ 3x=0 \\ \\ x=0 math

math \log_3{\big(\frac{1}{9}\big)}=x \;\; \iff \;\; 3^x=\frac{1}{9} math
 * (c) **

math \\ 3^x=\dfrac{1}{3^2} \\ \\ 3^x=3^{-2} \\ \\ x=-2 math

**Example 2**
Solve for **//x//**: math \log_x{27}=\frac{3}{2} math

__Solution:__

math \log_x{27}=\frac{3}{2} \iff x^{\frac{3}{2}}=27 math math x^{\frac{3}{2}}=3^3 math math \big(x^{\frac{1}{2}}\big)^3=3^3 math math x^{\frac{1}{2}}=3 math square both sides math x=9 math

**Example 3**
Use the calculator to find x correct to 3 decimal places. math \log_x5=3 math

__**Solution:**__

x = 1.710

**Example 4**
math \text{Solve for x:}\quad \log_2{(x-1)} + \log_2{(x+2)} = \log_2{(6x-8)} math

__Solution:__

math \log_2{(x-1)}+ \log_2{(x+2)}= \log_2{(6x-8)} math

Use log laws to combine left side of the equation math \log_2{(x-1)(x+2)}= \log_2{(6x-8)} math

We have log(a) = log(b) so a = b math \\ (x-1)(x+2)=(6x-8) \\ \\ x^2+x-2=6x-8 \\ \\ x^2-5x+6=0 \\ \\ (x-3)(x-2)=0 \\ \\ math hence math x=2 \;\; \textit{ or } \;\; x = 3 math

Can we accept both solutions?

log(y) is only defined when y > 0

Go back to the original question. For each value of x the term inside each log must be positive: Is x – 1 > 0? Is x + 2 > 0 ? Is 6x – 8 > 0?

In this case, this is true for 2 and 3 so __accept__ both answers. Go to top of page flat .