Question 170138
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Table of Contents:

<a href="#90">Problem #90</a>
<a href="#94">Problem #94</a>




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90)


{{{3^(2x)-8*3^(x)+15=0}}} Start with the given equation



Let {{{z=3^(x)}}}. So this means that {{{z^2=(3^(x))^2=3^(2x)}}} (ie {{{z=3^(2x)}}})



{{{z^2-8z+15=0}}} Replace {{{3^(2x)}}} with {{{z^2}}} and {{{3^(x)}}} with "z" (see above)



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=1}}}, {{{b=-8}}}, and {{{c=15}}}



Let's use the quadratic formula to solve for z



{{{z = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{z = (-(-8) +- sqrt( (-8)^2-4(1)(15) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-8}}}, and {{{c=15}}}



{{{z = (8 +- sqrt( (-8)^2-4(1)(15) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{z = (8 +- sqrt( 64-4(1)(15) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{z = (8 +- sqrt( 64-60 ))/(2(1))}}} Multiply {{{4(1)(15)}}} to get {{{60}}}



{{{z = (8 +- sqrt( 4 ))/(2(1))}}} Subtract {{{60}}} from {{{64}}} to get {{{4}}}



{{{z = (8 +- sqrt( 4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (8 +- 2)/(2)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{z = (8 + 2)/(2)}}} or {{{z = (8 - 2)/(2)}}} Break up the expression. 



{{{z = (10)/(2)}}} or {{{z =  (6)/(2)}}} Combine like terms. 



{{{z = 5}}} or {{{z = 3}}} Simplify. 



Now remember, we let {{{z=3^(x)}}}. So this means that {{{3^(x) = 5}}} or {{{3^(x) = 3}}}



{{{3^(x) = 5}}} Start with the first equation



{{{x = log(3,(5))}}} Take the log base 3 of both sides (to isolate "x")


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{{{3^(x) = 3}}} Move onto the second equation



{{{x = log(3,(3))}}} Take the log base 3 of both sides (to isolate "x")



{{{x = 1}}} Evaluate the log base 3 of 3 to get 1



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Answer:



So the solutions are {{{x = log(3,(5))}}} or {{{x =1}}} 





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94)



Is the equation {{{x=(log(125,(5)))^(log(5,(125)))}}}????



If so, then....



{{{x=(log(125,(5)))^(log(5,(125)))}}} Start with the given equation



Let {{{w=log(125,(5))}}} and {{{z=log(5,(125))}}}



So the equation then becomes...



{{{x=(w)^(z)}}} 



Now let's evaluate the individual pieces:



{{{w=log(125,(5))}}} Start with the first assignment



{{{w=log(10,(5))/log(10,(125))}}} Use the change of base formula to break up the log



{{{w=log(10,(5))/log(10,(5^3))}}} Rewrite 125 as {{{5^3}}}. Note: {{{5^3=125}}}



{{{w=log(10,(5))/(3*log(10,(5)))}}} Pull the exponent down and place it out front.



{{{w=1/3}}} Divide and cancel out the common terms.



Since {{{w=1/3}}}, this means that {{{log(125,(5))=1/3}}} (this is the first piece)



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{{{z=log(5,(125))}}} Now move onto the second assignment



{{{z=log(10,(125))/log(10,(5))}}} Use the change of base formula to break up the log



{{{z=log(10,(5^3))/log(10,(5))}}} Rewrite 125 as {{{5^3}}}. Note: {{{5^3=125}}}



{{{z=(3*log(10,(5)))/log(10,(5))}}} Pull down the exponent.



{{{z=3}}} Divide and cancel out the common terms.



Since {{{z=3}}}, this means that {{{log(5,(125))=3}}} (which is the second piece)




{{{x=w^z}}} Go back to the original equation



{{{x=(1/3)^3}}} Plug in {{{w=1/3}}} and {{{z=3}}}



{{{x=(1^3)/(3^3)}}} Distribute the exponent.



{{{x=1/27}}} Cube 1 to get 1. Cube 3 to get 27. 




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Answer:


So the answer is {{{x=1/27}}}



This means that {{{(log(125,(5)))^(log(5,(125)))=1/27}}}