MHB Solving for an exponential equation using logarithms 16^{x}-5(4)^{x}-6=0

AI Thread Summary
The discussion revolves around solving the exponential equation 16^x - 5(4)^x - 6 = 0. The original poster struggles with applying logarithmic properties correctly and is confused about arriving at the solution log4(6). Another participant suggests rewriting the equation as a quadratic in terms of 4^x, which simplifies the problem. This approach allows the use of the quadratic formula to find the solution. The conversation concludes with the poster expressing gratitude for the clarification.
sp3
Messages
8
Reaction score
0
Hello I'm having trouble solving for this exponential equation : 16^{x}-(5,4)^{x}-6=0
I used some logarithms properties but can't get anything close to the following solutions here View attachment 8366
I tried using log base 16 : log16(16^{x})-6=log16((5,4)^{x}) ; then x - xlog16(5,4)=6 ;
factorizing x : x(1-log16(5,4))=6 here I get lost... I don't know how they got to log base 4 ( the answer is log4(6)) ... i thought about rewriting 5,4 as a fraction 27/5 but it doesn't help a lot... thanks in advance for the help
 

Attachments

  • solutions45.png
    solutions45.png
    5.8 KB · Views: 164
Mathematics news on Phys.org
Are you sure you've copied the equation correctly? According to W|A, the solution to the given equation is

$$x=0.833215$$

But:

$$\log_4(6)\approx1.292481250360578$$
 
Hi sp3, welcome to MHB! ;)

Can it be that your equation should be $16^x-5\cdot 4^x-6=0$?

If so then we can write it as $(4^x)^2-5(4^x)-6=0$ and apply the quadratic formula.
 
Thank youu I suspected something was up with this decimal number... thanks a million times guys! :D
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top