To solve the equation [tex]\(\log_2(x) = \log_{12}(x)\)[/tex] by graphing, we need to graph the appropriate logarithmic functions and find their intersection point.
### Step-by-Step Solution:
1. Identify the equations to graph:
We know the given equation is [tex]\(\log_2(x) = \log_{12}(x)\)[/tex].
- For [tex]\(\log_2(x)\)[/tex], we can use the change of base formula:
[tex]\[
y_1 = \log_2(x) = \frac{\log(x)}{\log(2)}
\][/tex]
- Similarly, for [tex]\(\log_{12}(x)\)[/tex], we use the change of base formula:
[tex]\[
y_2 = \log_{12}(x) = \frac{\log(x)}{\log(12)}
\][/tex]
So, the two functions we need to graph are:
[tex]\[
y_1 = \frac{\log(x)}{\log(2)}
\][/tex]
[tex]\[
y_2 = \frac{\log(x)}{\log(12)}
\][/tex]
2. Graph the functions:
- The function [tex]\(y_1 = \frac{\log(x)}{\log(2)}\)[/tex] represents the logarithm of [tex]\(x\)[/tex] with base 2.
- The function [tex]\(y_2 = \frac{\log(x)}{\log(12)}\)[/tex] represents the logarithm of [tex]\(x\)[/tex] with base 12.
3. Find the intersection point:
By plotting these two functions on the same graph, we can identify the [tex]\(x\)[/tex]-value where they intersect, which is the solution to the equation [tex]\(\log_2(x) = \log_{12}(x)\)[/tex].
### Interpretation:
[tex]\(y_1 = \frac{\log(x)}{\log(2)}\)[/tex] will have a steeper curve compared to [tex]\(y_2 = \frac{\log(x)}{\log(12)}\)[/tex], because the base 2 logarithm increases more rapidly than the base 12 logarithm.
### Graphical Solution:
The point where the two graphs intersect occurs when [tex]\(y_1 = y_2\)[/tex]:
[tex]\[
\frac{\log(x)}{\log(2)} = \frac{\log(x)}{\log(12)}
\][/tex]
Canceling the [tex]\(\log(x)\)[/tex] on both sides (since [tex]\(\log(x) \neq 0\)[/tex]):
[tex]\[
\frac{1}{\log(2)} = \frac{1}{\log(12)}
\][/tex]
This simplifies to:
[tex]\[
\log(2) = \log(12)
\][/tex]
However, that's incorrect since [tex]\(\log\)[/tex] is the same function for different bases and doesn't cancel that way. Instead, simplifying it algebraically:
[tex]\[
x^{\log(12)}=x^{\log(2)}
\][/tex]
For those equal x must be 1
Therefore:
[tex]\[
x = 1
\][/tex]
The graphs of the equations [tex]\(y_1 = \log_2(x)\)[/tex] and [tex]\(y_2 = \log_{12}(x)\)[/tex] intersect at [tex]\(x = 1\)[/tex].
### Conclusion:
From graphing these functions, the solution to the equation [tex]\(\log_2(x) = \log_{12}(x)\)[/tex] is:
[tex]\[
x = 1
\][/tex]
