Question 173334
# 1


In order to graph {{{f(x)=7^x}}}, we need to plot some points:



Note: I'll use "y" in place of f(x)




Let's find the y value when {{{x=-2}}}



{{{y=7^x}}} Start with the given equation



{{{y=7^(-2)}}} Plug in {{{x=-2}}}



{{{y=(1/7)^2}}} Flip the fraction to make the exponent positive.



{{{y=1/49}}} Square {{{1/7}}} to get {{{1/49}}}



{{{y=0.02041}}} Divide



So when {{{x=-2}}}, then {{{y=0.02041}}}



------------------------------------------------------------------



Let's find the y value when {{{x=-1}}}



{{{y=7^x}}} Start with the given equation



{{{y=7^(-1)}}} Plug in {{{x=-1}}}



{{{y=(1/7)^1}}} Flip the fraction to make the exponent positive.



{{{y=1/7}}} Raise {{{1/7}}} to the first power to get {{{1/7}}}



{{{y=0.14286}}} Divide



So when {{{x=-1}}}, then {{{y=0.14286}}}



------------------------------------------------------------------



Let's find the y value when {{{x=0}}}



{{{y=7^x}}} Start with the given equation



{{{y=7^(0)}}} Plug in {{{x=0}}}



{{{y=1}}} Raise {{{7}}} to the zeroth power to get {{{1}}}




So when {{{x=0}}}, then {{{y=1}}}



------------------------------------------------------------------



Let's find the y value when {{{x=1}}}



{{{y=7^x}}} Start with the given equation



{{{y=7^(1)}}} Plug in {{{x=1}}}



{{{y=7}}} Raise {{{7}}} to the first power to get {{{7}}}




So when {{{x=1}}}, then {{{y=7}}}



------------------------------------------------------------------



So we have the following values:



<table width="100" border="1"><th>x</th><th>y</th>
<tr><td>-2</td><td>0.02041</td></tr>
<tr><td>-1</td><td>0.14286</td></tr>
<tr><td>0</td><td>1</td></tr>
<tr><td>1</td><td>7</td></tr>
</table>




Now let's plot these points



{{{ drawing(500, 500, -5, 5, -2, 10,
 grid(1),
 graph( 500, 500, -5, 5, -2, 10,0),
 circle(-2,0.02041,0.02),
 circle(-2,0.02041,0.04),
 circle(-2,0.02041,0.06),
 circle(-1,0.14286,0.02),
 circle(-1,0.14286,0.04),
 circle(-1,0.14286,0.06),
 circle(0,1,0.02),
 circle(0,1,0.04),
 circle(0,1,0.06),
 circle(1,7,0.02),
 circle(1,7,0.04),
 circle(1,7,0.06)

)}}}



Now draw a curve through those points to graph {{{f(x)=(1/5)^x}}}



{{{ drawing(500, 500, -5, 5, -2, 10,
 grid(1),
 graph( 500, 500, -5, 5, -2, 10,(7)^x),
 circle(-2,0.02041,0.02),
 circle(-2,0.02041,0.04),
 circle(-2,0.02041,0.06),
 circle(-1,0.14286,0.02),
 circle(-1,0.14286,0.04),
 circle(-1,0.14286,0.06),
 circle(0,1,0.02),
 circle(0,1,0.04),
 circle(0,1,0.06),
 circle(1,7,0.02),
 circle(1,7,0.04),
 circle(1,7,0.06)

)}}}



Note: the graph does NOT cross or touch the x-axis. It just looks like that since the y values become very small.



<hr>




# 2




In order to graph {{{f(x)=log(3,(x))}}}, we need to plot some points:



Note: I'll use "y" in place of f(x)



Let's find the y value when {{{x=1}}}



{{{y=log(3,(x))}}} Start with the given equation



{{{y=log(3,(1))}}} Plug in {{{x=1}}}



{{{y=0}}} Use a calculator to evaluate log base 3 of 1 to get 0



So when {{{x=1}}}, then {{{y=0}}}



-----------------------------------------------




Let's find the y value when {{{x=2}}}



{{{y=log(3,(x))}}} Start with the given equation



{{{y=log(3,(2))}}} Plug in {{{x=2}}}



{{{y=0.63093}}} Use a calculator to evaluate log base 3 of 2 to get 0.63093



So when {{{x=2}}}, then {{{y=0.63093}}}



-----------------------------------------------




Let's find the y value when {{{x=3}}}



{{{y=log(3,(x))}}} Start with the given equation



{{{y=log(3,(3))}}} Plug in {{{x=3}}}



{{{y=1}}} Use a calculator to evaluate log base 3 of 3 to get 1



So when {{{x=3}}}, then {{{y=1}}}



-----------------------------------------------




Let's find the y value when {{{x=4}}}



{{{y=log(3,(x))}}} Start with the given equation



{{{y=log(3,(4))}}} Plug in {{{x=4}}}



{{{y=1.26186}}} Use a calculator to evaluate log base 3 of 4 to get 1.26186



So when {{{x=4}}}, then {{{y=1.26186}}}



-----------------------------------------------



Let's find the y value when {{{x=5}}}



{{{y=log(3,(x))}}} Start with the given equation



{{{y=log(3,(5))}}} Plug in {{{x=5}}}



{{{y=1.26186}}} Use a calculator to evaluate log base 3 of 5 to get 1.46497



So when {{{x=5}}}, then {{{y=1.46497}}}



-----------------------------------------------





So we have the following values:



<table width="100" border="1"><th>x</th><th>y</th>
<tr><td>1</td><td>0</td></tr>
<tr><td>2</td><td>0.63093</td></tr>
<tr><td>3</td><td>1</td></tr>
<tr><td>4</td><td>1.26186</td></tr>
<tr><td>5</td><td>1.46497</td></tr>
</table>




Now let's plot these points



{{{ drawing(500, 500, -10, 10, -10, 10,
 grid(1),
 graph( 500, 500, -10, 10, -10, 10,0),
 circle(1,0,0.05),
 circle(1,0,0.08),
 circle(1,0,0.1),
 circle(2,0.63093,0.05),
 circle(2,0.63093,0.08),
 circle(2,0.63093,0.1),
 circle(3,1,0.05),
 circle(3,1,0.08),
 circle(3,1,0.1),
 circle(4,1.26186,0.05),
 circle(4,1.26186,0.08),
 circle(4,1.26186,0.1),
 circle(5,1.46497,0.05),
 circle(5,1.46497,0.08),
 circle(5,1.46497,0.1)

)}}}



Now draw a curve through those points to graph {{{f(x)=log(3,(x))}}}



{{{ drawing(500, 500, -10, 10, -10, 10,
 grid(1),
 graph( 500, 500, -10, 10, -10, 10,log(3,(x))),
 circle(1,0,0.05),
 circle(1,0,0.08),
 circle(1,0,0.1),
 circle(2,0.63093,0.05),
 circle(2,0.63093,0.08),
 circle(2,0.63093,0.1),
 circle(3,1,0.05),
 circle(3,1,0.08),
 circle(3,1,0.1),
 circle(4,1.26186,0.05),
 circle(4,1.26186,0.08),
 circle(4,1.26186,0.1),
 circle(5,1.46497,0.05),
 circle(5,1.46497,0.08),
 circle(5,1.46497,0.1)

)}}}