Question

5^(-x)+7=2x+4 This was on plato

Answer 1

Below

I hope its not too complicated

`x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{(3\ln \left(5\right))/(2)}}\right)}{\ln \left(5\right)}+(3)/(2)`

Explanation:

`5^(\left(-x\right))+7=2x+4\\\\\mathrm{Prepare}\:5^(\left(-x\right))+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^(\ln \left(5\right)x)\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-(3)/(2)\right)\ln \left(5\right)=u\mathrm{\:and\:}x=(u)/(\ln \left(5\right))+(3)/(2)\\\\1=\left(2\left((u)/(\ln \left(5\right))+(3)/(2)\right)-3\right)e^{\ln \left(5\right)\left((u)/(\ln \left(5\right))+(3)/(2)\right)}`

`Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+(3)/(2)\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{(2u+3\ln \left(5\right))/(2)}u}{e^{(3\ln \left(5\right))/(2)}}=\frac{\ln \left(5\right)}{2e^{(3\ln \left(5\right))/(2)}}`

`\mathrm{Solve\:}\:\frac{e^{(2u+3\ln \left(5\right))/(2)}u}{e^{(3\ln \left(5\right))/(2)}}=\frac{\ln \left(5\right)}{2e^{(3\ln \left(5\right))/(2)}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{(3\ln \left(5\right))/(2)}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-(3)/(2)\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x`

`\mathrm{Solve\:}\:\left(x-(3)/(2)\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{(3\ln \left(5\right))/(2)}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{(3\ln \left(5\right))/(2)}}\right)}{\ln \left(5\right)}+(3)/(2)`