Solve for $x$.
Im having trouble. I solved a similar problem $2^{x} + x = 37$, but ive been having trouble with this one. Could anyone point me in the right direction? Thanks
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1Where do you want your solutions to lie? – ShyamalSayak Jan 28 '24 at 04:39
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You're not going to find a solution without the Lambert $W$ function. Luckily I wrote a more general question addressing that here – PrincessEev Jan 28 '24 at 04:50
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1Does this answer your question? How does one solve for $x$ in the equality $a^x = bx + c$? – PrincessEev Jan 28 '24 at 04:50
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Do you need to solve for intergal $x$ or real $x$ (in which case you will have to use Lambert W) – Sahaj Jan 28 '24 at 05:09
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1Can you tell us how you solved $2^x+x=37$ and where that method failed for this equation? – ultralegend5385 Jan 28 '24 at 05:28
1 Answers
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As mentioned elsewhere, you could use the Lambert W function after a lot of algebra. You can get an answer by figuring out what integers straddle the Real solution, then derive a new equation for an "offset" from the integer guess. Using Taylor series you can translate that equation into a linear equation solvable with simple algebra techniques.
$3^x-2x=25$
$f(x)=3^x-2x-25$
$f'(x)=\ln3 \cdot 3^x-2$
$f'(x)=0\implies x=\ln(\frac{2}{\ln 3})/\ln 3 \approx 0.54$
$f(0)=-24$ ,$f(1)=-24$,$f(2)=-20$
$f(3)=-4$,$f(4)=48$
So the answer is between $3$ and $4$.
$3^x=2x+25$
$x=3+\delta$
$3^{3+\delta}=2(3+\delta)+25$
$27\cdot 3^\delta=2\delta + 31$
$3\ln 3 + \delta \ln 3=\ln 31+\ln(1+2\delta/31)$
$\frac{1}{1+x}=1-x+x^2-x^3+...$
$\ln|1+x|=x-x^2/2+...$
$3\ln3+\delta \ln 3 = \ln 31 + 2\delta/31$
$\frac{3\ln3-\ln 31}{2/31-\ln 3}=\delta$
$3+\delta = \frac{6/31-\ln31}{2/31-\ln 3}\approx 3.13359525339$
If $c=3.13359525339$, then $3^c-2c-25\approx 0.0011$
TurlocTheRed
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