Question 156683
<a name="one">


<a href="#two">Jump to transformations of the logarithmic function</a>


# 1


First let's calculate the values for {{{y=e^x}}}



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



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



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



{{{y=0.135}}} Use a calculator to evaluate {{{e^(-2)}}} to get {{{0.135}}}



So if {{{x=-2}}}, then {{{y=0.135}}}. So we have the point (-2,0.135)



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



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



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



{{{y=0.368}}} Use a calculator to evaluate {{{e^(-1)}}} to get {{{0.368}}}



So if {{{x=-1}}}, then {{{y=0.368}}}. So we have the point (-1,0.368)



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



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



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



{{{y=1}}} Use a calculator to evaluate {{{e^(0)}}} to get {{{1}}}



So if {{{x=0}}}, then {{{y=1}}}. So we have the point (0,1)



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



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



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



{{{y=2.718}}} Use a calculator to evaluate {{{e^(1)}}} to get {{{2.718}}}



So if {{{x=1}}}, then {{{y=2.718}}}. So we have the point (1,2.718)



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



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



{{{y=e^(2)}}} Plug in {{{x=2}}}



{{{y=7.389}}} Use a calculator to evaluate {{{e^(2)}}} to get {{{7.389}}}



So if {{{x=2}}}, then {{{y=7.389}}}. So we have the point (2,7.389)




Now let's plot these points


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/ziggy%20rayon/solution2apoints9-10-08.jpg" alt="Photobucket - Video and Image Hosting">



Now draw a line through the points to graph {{{y=e^x}}}



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/ziggy%20rayon/solution2agraph9-10-08.jpg" alt="Photobucket - Video and Image Hosting">

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# 2


Now let's calculate the values for {{{y=e^(2x)}}}



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



{{{y=e^(2x)}}} Start with the given equation



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



{{{y=e^(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}



{{{y=0.018}}} Use a calculator to evaluate {{{e^(-4)}}} to get {{{0.018}}}



So if {{{x=-2}}}, then {{{y=0.018}}}. So we have the point (-2,0.018)



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



{{{y=e^(2x)}}} Start with the given equation



{{{y=e^(2(-1))}}} Plug in {{{x=-1}}}



{{{y=e^(-2)}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}



{{{y=0.135}}} Use a calculator to evaluate {{{e^(-2)}}} to get {{{0.135}}}



So if {{{x=-1}}}, then {{{y=0.135}}}. So we have the point (-1,0.135)



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



{{{y=e^(2x)}}} Start with the given equation



{{{y=e^(2(0))}}} Plug in {{{x=0}}}



{{{y=e^(0)}}} Multiply {{{2}}} and {{{0}}} to get {{{0}}}



{{{y=1}}} Use a calculator to evaluate {{{e^(0)}}} to get {{{1}}}



So if {{{x=0}}}, then {{{y=1}}}. So we have the point (0,1)



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



{{{y=e^(2x)}}} Start with the given equation



{{{y=e^(2(1))}}} Plug in {{{x=1}}}



{{{y=e^(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}



{{{y=7.389}}} Use a calculator to evaluate {{{e^(2)}}} to get {{{7.389}}}



So if {{{x=1}}}, then {{{y=7.389}}}. So we have the point (1,7.389)



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



{{{y=e^(2x)}}} Start with the given equation



{{{y=e^(2(2))}}} Plug in {{{x=2}}}



{{{y=e^(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}



{{{y=54.598}}} Use a calculator to evaluate {{{e^(4)}}} to get {{{54.598}}}



So if {{{x=2}}}, then {{{y=54.598}}}. So we have the point (2,54.598)



Now let's plot these points


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/ziggy%20rayon/solution2bpoints9-10-08.jpg" alt="Photobucket - Video and Image Hosting">



Now draw a line through the points to graph {{{y=e^(2x)}}}



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/ziggy%20rayon/solution2bgraph9-10-08.jpg" alt="Photobucket - Video and Image Hosting">



Notice how the graph of {{{y=e^(2x)}}} is much steeper than the graph of {{{y=e^(x)}}}. This is because the values of {{{y=e^(2x)}}} are simply the values of {{{y=e^(x)}}} squared (since {{{e^(2x)=(e^(x))^2}}}). For instance, if {{{x=1}}}, then {{{y=e^1=e}}} and {{{y=e^(2*1)=e^2}}}. Since {{{e^2}}} is the square of {{{e}}}, this verifies our claim. So the transformation that occurs on {{{y=e^(2x)}}} is a horizontal compression (since we are dealing with "x" and the graph gets thinner).


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<a name="two">


<a href="#one">Jump to transformations of the exponential function</a>

# 3



Now let's calculate the values for {{{y=log(10,(x))}}}



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



{{{y=log(10,(x))}}} Start with the given equation



{{{y=log(10,(1))}}} Plug in {{{x=1}}}



{{{y=0}}} Use a calculator to evaluate {{{log(10,(1))}}} to get {{{0}}}



So if {{{x=1}}}, then {{{y=0}}}. So we have the point (1,0)


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



{{{y=log(10,(x))}}} Start with the given equation



{{{y=log(10,(2))}}} Plug in {{{x=2}}}



{{{y=0.301}}} Use a calculator to evaluate {{{log(10,(2))}}} to get {{{0.301}}}



So if {{{x=2}}}, then {{{y=0.301}}}. So we have the point (2,0.301)


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



{{{y=log(10,(x))}}} Start with the given equation



{{{y=log(10,(3))}}} Plug in {{{x=3}}}



{{{y=0.477}}} Use a calculator to evaluate {{{log(10,(3))}}} to get {{{0.477}}}



So if {{{x=3}}}, then {{{y=0.477}}}. So we have the point (3,0.477)


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



{{{y=log(10,(x))}}} Start with the given equation



{{{y=log(10,(4))}}} Plug in {{{x=4}}}



{{{y=0.602}}} Use a calculator to evaluate {{{log(10,(4))}}} to get {{{0.602}}}



So if {{{x=4}}}, then {{{y=0.602}}}. So we have the point (4,0.602)


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



{{{y=log(10,(x))}}} Start with the given equation



{{{y=log(10,(5))}}} Plug in {{{x=5}}}



{{{y=0.699}}} Use a calculator to evaluate {{{log(10,(5))}}} to get {{{0.699}}}



So if {{{x=5}}}, then {{{y=0.699}}}. So we have the point (5,0.699)




Now let's plot the points


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/solution2cpointsedit9-10-08.jpg" alt="Photobucket - Video and Image Hosting">



Now connect the points to graph {{{y=log(10,(x))}}}

<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/solution2cgraphedit9-10-08.jpg" alt="Photobucket - Video and Image Hosting">



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# 4



Now let's calculate the values for {{{y=log(10,(2x))}}}



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



{{{y=log(10,(2x))}}} Start with the given equation



{{{y=log(10,(2*1))}}} Plug in {{{x=1}}}



{{{y=log(10,(2))}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}



{{{y=0.301}}} Use a calculator to evaluate {{{log(10,(2))}}} to get {{{0.301}}}



So if {{{x=1}}}, then {{{y=0.301}}}. So we have the point (1,0.301)


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



{{{y=log(10,(2x))}}} Start with the given equation



{{{y=log(10,(2*2))}}} Plug in {{{x=2}}}



{{{y=log(10,(4))}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}



{{{y=0.602}}} Use a calculator to evaluate {{{log(10,(4))}}} to get {{{0.602}}}



So if {{{x=2}}}, then {{{y=0.602}}}. So we have the point (2,0.602)


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



{{{y=log(10,(2x))}}} Start with the given equation



{{{y=log(10,(2*3))}}} Plug in {{{x=3}}}



{{{y=log(10,(6))}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}



{{{y=0.778}}} Use a calculator to evaluate {{{log(10,(6))}}} to get {{{0.778}}}



So if {{{x=3}}}, then {{{y=0.778}}}. So we have the point (3,0.778)


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



{{{y=log(10,(2x))}}} Start with the given equation



{{{y=log(10,(2*4))}}} Plug in {{{x=4}}}



{{{y=log(10,(8))}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}



{{{y=0.903}}} Use a calculator to evaluate {{{log(10,(8))}}} to get {{{0.903}}}



So if {{{x=4}}}, then {{{y=0.903}}}. So we have the point (4,0.903)


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



{{{y=log(10,(2x))}}} Start with the given equation



{{{y=log(10,(2*5))}}} Plug in {{{x=5}}}



{{{y=log(10,(10))}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}



{{{y=1}}} Use a calculator to evaluate {{{log(10,(10))}}} to get {{{1}}}



So if {{{x=5}}}, then {{{y=1}}}. So we have the point (5,1)




Now let's plot the points


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/solution2dpointsedit9-10-08.jpg" alt="Photobucket - Video and Image Hosting">



Now draw a curve through the points to graph {{{y=log(10,(2x))}}}


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/solution2dgraphedit9-10-08.jpg" alt="Photobucket - Video and Image Hosting">





By comparing {{{y=log(10,(x))}}} and {{{y=log(10,(2x))}}}, we can see that the graph of {{{y=log(10,(2x))}}} is simply the graph of {{{y=log(10,(x))}}} shifted up some number of units. So a vertical translation has occured on {{{y=log(10,(x))}}} to get {{{y=log(10,(2x))}}}