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

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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
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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
 

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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
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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