Finding the root of a Transcendental equation [closed]

I'm trying to find R in the transcendental equation.

$$R + u_r\,t=(\frac{a_1\,B}{\nu}\, \gamma_0^2)^{\frac{3}{4}} \sigma R \,e^{\frac{3 \sigma}{v_{sh}\,t_{acc}}(R-R^*)}$$

Everything except R is a constant i.e.

$$R + K_1=K_2 R \,e^{K_3(R-K_4)}$$

where $$K_{1,2,3,4}$$ are known constants.

How can I use Mathematica for find R?

closed as off-topic by Michael E2, Henrik Schumacher, march, bbgodfrey, Vitaliy KaurovDec 4 '18 at 19:00

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• If all constants have numerical values, use FindRoot. – bbgodfrey Nov 25 '18 at 14:05

K1 = 1; K2 = 1; K3 = 1; K4 = 1;

Plot[r + K1 - K2*r*Exp[K3 (r - K4)], {r, -2, 2}]


For exact solutions (expressed as Root objects) use Solve by constraining the range of r

Solve[{r + K1 == K2*r*Exp[K3 (r - K4)], -2 < r < 2}, r]

(* {{r -> Root[{-1 - #1 + E^(-1 + #1) #1 &, -1.13422002963071713234}]}, {r ->
Root[{-1 - #1 + E^(-1 + #1) #1 &, 1.50855472406037552015}]}} *)


Use N to get approximate numeric values.

% // N

(* {{r -> -1.13422}, {r -> 1.50855}} *)


Similarly with NSolve

NSolve[{r + K1 == K2*r*Exp[K3 (r - K4)], -2 < r < 2}, r]

(* {{r -> -1.13422}, {r -> 1.50855}} *)


With FindRoot you need to use two starting values

FindRoot[r + K1 == K2*r*Exp[K3 (r - K4)], {r, #}] & /@ {-2, 2}

(* {{r -> -1.13422}, {r -> 1.50855}} *)