Solve the given problem involving logarithms

[chwala]

Homework Statement

This is my own question (set by me).

Solve for ##x## given,

##\log_{4} x = \log_{11} (x+6)##.

Relevant Equations

understanding of change of base.

In my working i have,

##\dfrac{\log_{11} x }{\log_{11} 4}= \log_{11} (x+6)##

##\dfrac{\log_{11} x }{0.5781}= \log_{11} (x+6)##

##\log_{11} x = \log_{11} \left[(x+6)\right]^{0.5781}##

##x^{1.729} = x+ 6##

##x^{1.729} -x-6=0##

Having ##f(x) = x^{1.729} -x-6##

At this point i made use of Newton's method. i.e

##x_{n+1} =x_n - \dfrac{f(x)}{f'(x)}##

Letting ##x_0 = 3##,

##x_1 = 3.8127##

##x_2= 3.8127- \dfrac{0.3019}{3.5868} = 3.72854##

##x_3= 3.72854 - \dfrac{0.00320}{3.512807} = 3.7191##

##x_4= 3.7191 - \dfrac{-0.029919}{3.504475} = 3.727##

##x≅3.73##

There may be a better approach hence my post. Cheers.

 

[Hill]

There may be a better approach

Yes. Using midpoint:
##f(x)= \log_{11} (x+6)-\log_{4} x##
##x=3, f(x)=0.12383257##
##x=4, f(x)=-0.039747432##
##x=3.5, f(x)=0.035184142##
##x=3.75, f(x)=-0.003751076##
##x=3.625, f(x)=0.015322586##
##x=3.6875, f(x)=0.005690808##
##x=3.71875, f(x)=0.000946551##
##x=3.734375, f(x)=-0.001408039##
##x=3.7265625, f(x)=-0.000232195##

 

[OmCheeto]

Since I'm not a mathematician, I have no idea what my method is called, but it seems to converge on the answer a bit quicker than Newton's method.

##f(x)= \log_{11} (x+6)-\log_{4} x##

##x_1 = 3, f(x)= 0.1238325696##
##x_2 = 4, f(x)= -0.0397474322##

From here I find the slope and intercept of the above two points and solve for x when f(x) is set to zero.
I then make that my new x and repeat the process until f(x) = 0

##x_3 = 3.757015333, f(x)= -0.0047993244##
##x_4 = 3.723646939, f(x)= 0.0002073635##
##x_5 = 3.725028968, f(x)= -0.0000010455##
##x_6 = 3.725022035, f(x)= -0.0000000002##
##x_7 = 3.725022033, f(x)= 0.0000000000##

Hopefully a real mathematician will recognize it and give it a name.

In any event

x≅3.73

appears to be correct.

 

[PeroK]

Or, put it in a spreadsheet:

x	f(x)
3.725022033325​	-2.07E-14​
3.725022033326​	1.30E-13​

 

[Gavran]

Iterates $$ x _ 3 = 3,757015333 $$ $$ x _ 4 = 3,723646939 $$ $$ x _ 5 = 3,725028968 $$ $$ x _ 6 = 3,725022035 $$ $$ x _ 7 = 3,725022033 $$ for the function $$ f ( x ) = log _ { 11 } ( x + 6 ) – log _ { 4 } x $$ with initials $$ x _ 1 = 3 $$ $$ x _ 2 = 4 $$ are produced by a secant method. In this method iterates are produced by using the recurrence relation $$ x _ { n } = x _ { n – 1 } – f ( x _ { n - 1 } ) \frac { x _ { n - 1 } – x _ { n - 2 } } { f ( x _ { n – 1 } ) – f ( x _ { n – 2 } ) } $$ and the method can be thought of as an approximation of a tangent method.

 

[OmCheeto]

Thanks Gavran!
I see that upon inspection that my equation
$$ x_3 = x_2 - \frac {y_2} {m_2} $$
is equivalent to your equation
$$ x _ { n } = x _ { n – 1 } – f ( x _ { n - 1 } ) \frac { x _ { n - 1 } – x _ { n - 2 } } { f ( x _ { n – 1 } ) – f ( x _ { n – 2 } ) } $$

Can't say I remember hearing of the 'Secant Method', but I find this interesting; "...the secant method predates Newton's method by over 3000 years."
ref: wiki