3^2x=5(3^x)+36 whats the exact value of x?

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I let A=3x and then i got (A-9)(A-4) when i factored the quadratic. Then when I got a common base of 3 I solved one x value to be x=2 but Im not sure about the other one? 3^x=4 what is the common base? I used the graphing calculator instead and got x=1.26 is this correct? could i have done it another way? :rolleyes:
 

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  • #2
Zurtex
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A=3^x not 3x right?

I've not checked your algebra but assuming it is right:

[tex]3^x = 4[/tex]

[tex]x = \log_3 4[/tex]

or:

[tex]3^x = 4[/tex]

[tex]\ln \left( 3^x \right) = \ln (4)[/tex]

[tex]x \ln (3) = \ln(4)[/tex]

[tex]x = \frac{\ln(4)}{\ln(3)}[/tex]

Edit: You have made a mistake, check your pluses and minus when you factored again, there is only one real root.
 
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Thanks I got it :smile:
 
  • #4
dextercioby
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aisha said:
I let A=3x and then i got (A-9)(A-4) when i factored the quadratic. Then when I got a common base of 3 I solved one x value to be x=2 but Im not sure about the other one? 3^x=4 what is the common base? I used the graphing calculator instead and got x=1.26 is this correct? could i have done it another way? :rolleyes:
Final answer,...............................................YES.But the equation subject of this post will not appear in the solution to your problem,as the decomposition u found was wrong.It should have stated (A-9)(A+4) with only the "9" sollution adittable,since the exponetial of real numbers cannot be negative.
Anyway,for the equation u stated,here goes my say:
U got [itex] 3^x =4 [/itex].Apply logarithm in any base on both sides of the eq.Apply it in the base 3 to find:[itex] x=\log_{3} 4 [/itex].Apply in the "natural" basis to find:[itex] x=\frac{\ln 4}{\ln 3} [/itex].
Take the pocket intelligent calculator (the one with scientifical functions) and compute the ratio between the 2 natural logarithms and you"ll find "x" with a staggering approximation (actuatlly the calculator's number of possible decimals).

Equations like the one u encountered (the last one) are called transcendental equations and,in general,solutions are transcendental numbers,i.e. numbers like "e" and "pi".

Daniel.

PS.Edit:Zurtex,i hadn't seen yor post,as i would have cut mine to half,sice you dealt with the transcedental eq.
 
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dextercioby said:
Final answer,...............................................YES.
U have done magnificiently up until the last equation.
U got [itex] 3^x =4 [/itex].Apply logarithm in any base on both sides of the eq.Apply it in the base 3 to find:[itex] x=\log_{3} 4 [/itex].Apply in the "natural" basis to find:[itex] x=\frac{\ln 4}{\ln 3} [/itex].
Take the pocket intelligent calculator (the one with scientifical functions) and compute the ratio between the 2 natural logarithms and you"ll find "x" with a staggering approximation (actuatlly the calculator's number of possible decimals).

Equations like the one u encountered (the last one) are called transcendental equations and,in general,solutions are transcendental numbers,i.e. numbers like "e" and "pi".

Daniel.
Since one value was negative I think there is no solution for that one since when a positive base is raised to any exponent the result should be a positive value. We are not doing log this year, thanks for ur help!
 
  • #6
dextercioby
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aisha said:
Since one value was negative I think there is no solution for that one since when a positive base is raised to any exponent the result should be a positive value. We are not doing log this year, thanks for ur help!
I edited my post.I assumed that the decomposition u found was correct (and so did Zurtex),so i proceded with the unclarities regarding the last equation.
Now my post makes perfect sense.Almost... :tongue2:
 
  • #7
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I dont understand am I wrong?
 
  • #8
dextercioby
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aisha said:
I let A=3x and then i got (A-9)(A-4) when i factored the quadratic.
YES,but,thankfully u figured out the mistake.

Daniel.
 

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