3^x+x = √3
Solution with iteration: 3^x = \sqrt{3} - x \quad | \quad \ln{()} \\\\ x\ln{(3)} = \ln{ ( \sqrt{3} - x ) } \\\\ \textcolor[rgb]{1,0,0}{\boxed{x_{new} = \dfrac{ \ln{ ( \sqrt{3} - x_{old} ) } }{\ln{(3)} } } }\\\\ x_{old} \small{\text{ starts with 1} } }
nxnewxold1−0.28390849201120.63816431742−0.2839084920130.081682084590.6381643174240.456028697830.0816820845950.221868546960.4560286978360.375228231280.2218685469670.277755517970.3752282312880.340904111590.2777555179790.300495790540.34090411159100.326558588180.30049579054110.309834123490.32655858818120.320601456230.30983412349130.313683989120.32060145623140.318134145650.31368398912150.315273760810.31813414565160.317113334880.31527376081170.315930692560.31711333488180.316691176900.31593069256190.316202229240.31669117690200.316516624590.31651662459210.316314479560.31651662459220.316444456780.31631447956230.316360884860.31644445678240.316414620280.31636088486250.316380069630.31641462028260.316402285060.31638006963270.316388001010.31640228506280.316397185380.31638800101290.316391280010.31639718538300.316395077050.31639128001310.316392635630.31639507705320.316394205410.31639263563330.316393196080.31639420541340.316393845060.31639319608350.316393427780.31639384506360.316393696080.31639342778370.316393523570.31639369608380.316393634490.31639352357390.316393563170.31639363449400.316393609030.31639356317410.316393579540.31639360903420.316393598500.31639357954430.316393586310.31639359850440.316393594150.31639358631450.316393589110.31639359415460.316393592350.31639359235470.316393590260.31639359235480.316393591600.31639359026490.316393590740.31639359160500.316393591300.31639359130510.316393590940.31639359130520.316393591170.31639359094530.316393591020.31639359117540.316393591120.31639359102550.316393591060.31639359112560.316393591090.31639359106570.316393591070.31639359109580.316393591090.31639359107590.316393591080.31639359109600.316393591080.31639359108.........
x=0.31639359108
f(x)=ln(√3−x)ln(3)g(x)=x We set f(x)=g(x) and iterate the cut between line and our function.
The convergence is given if the derivation |f′(xStart)|<1
f(x)=ln(√3−x)ln(3)f′(x)=1ln(3)∗(−1)(√3−x)
The graphical soln shows the real solution to be approx x=0.316
Here is the graph
https://www.desmos.com/calculator/msrf4kimfo
BUT I do not know how to do this algebraically.
3^x+x = √3
Solution with iteration: 3^x = \sqrt{3} - x \quad | \quad \ln{()} \\\\ x\ln{(3)} = \ln{ ( \sqrt{3} - x ) } \\\\ \textcolor[rgb]{1,0,0}{\boxed{x_{new} = \dfrac{ \ln{ ( \sqrt{3} - x_{old} ) } }{\ln{(3)} } } }\\\\ x_{old} \small{\text{ starts with 1} } }
nxnewxold1−0.28390849201120.63816431742−0.2839084920130.081682084590.6381643174240.456028697830.0816820845950.221868546960.4560286978360.375228231280.2218685469670.277755517970.3752282312880.340904111590.2777555179790.300495790540.34090411159100.326558588180.30049579054110.309834123490.32655858818120.320601456230.30983412349130.313683989120.32060145623140.318134145650.31368398912150.315273760810.31813414565160.317113334880.31527376081170.315930692560.31711333488180.316691176900.31593069256190.316202229240.31669117690200.316516624590.31651662459210.316314479560.31651662459220.316444456780.31631447956230.316360884860.31644445678240.316414620280.31636088486250.316380069630.31641462028260.316402285060.31638006963270.316388001010.31640228506280.316397185380.31638800101290.316391280010.31639718538300.316395077050.31639128001310.316392635630.31639507705320.316394205410.31639263563330.316393196080.31639420541340.316393845060.31639319608350.316393427780.31639384506360.316393696080.31639342778370.316393523570.31639369608380.316393634490.31639352357390.316393563170.31639363449400.316393609030.31639356317410.316393579540.31639360903420.316393598500.31639357954430.316393586310.31639359850440.316393594150.31639358631450.316393589110.31639359415460.316393592350.31639359235470.316393590260.31639359235480.316393591600.31639359026490.316393590740.31639359160500.316393591300.31639359130510.316393590940.31639359130520.316393591170.31639359094530.316393591020.31639359117540.316393591120.31639359102550.316393591060.31639359112560.316393591090.31639359106570.316393591070.31639359109580.316393591090.31639359107590.316393591080.31639359109600.316393591080.31639359108.........
x=0.31639359108
f(x)=ln(√3−x)ln(3)g(x)=x We set f(x)=g(x) and iterate the cut between line and our function.
The convergence is given if the derivation |f′(xStart)|<1
f(x)=ln(√3−x)ln(3)f′(x)=1ln(3)∗(−1)(√3−x)