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

[sp3]

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

 

[MarkFL]

Are you sure you've copied the equation correctly? According to W|A, the solution to the given equation is

\(\displaystyle x=0.833215\)

But:

\(\displaystyle \log_4(6)\approx1.292481250360578\)

 

[I like Serena]

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.

 

[sp3]

Thank youu I suspected something was up with this decimal number... thanks a million times guys! :D