Solve for x in the following equation: e^x + 10e^(-x) = 7
First of all, we bring the 7 to the left side of the equation to get: ex-7+10e-x=0. Then, by multiplying both sides of the equation by ex, we can get an equation in the form of a quadratic equation: e2x-7ex+10=0. By setting y = ex, the quadratic nature of the equation can be seen as it simplifies to y2-7y+10=0. From GCSE maths, we know this can be factorised to obtain (y-5)(y-2)=0 and see that y=2 or y=5. The final step for this question is to sub ex back into the equation and solve for x using the ln laws: ex = 2 or 5; therefore x = ln2 or ln5.
LB
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