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

[aisha]

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:

[Zurtex]

A=3^x not 3x right?

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

3^x = 4

x = \log_3 4

or:

3^x = 4

\ln \left( 3^x \right) = \ln (4)

x \ln (3) = \ln(4)

x = \frac{\ln(4)}{\ln(3)}

Edit: You have made a mistake, check your pluses and minus when you factored again, there is only one real root.

[aisha]

Thanks I got it :smile:

[dextercioby]

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 3^x =4 .Apply logarithm in any base on both sides of the eq.Apply it in the base 3 to find: x=\log_{3} 4 .Apply in the "natural" basis to find: x=\frac{\ln 4}{\ln 3} .
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.

[aisha]

Final answer,........................................... ....YES.
U have done magnificiently up until the last equation.
U got 3^x =4 .Apply logarithm in any base on both sides of the eq.Apply it in the base 3 to find: x=\log_{3} 4 .Apply in the "natural" basis to find: x=\frac{\ln 4}{\ln 3} .
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!

[dextercioby]

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:

[aisha]

I dont understand am I wrong?

[dextercioby]

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.