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: e^x-7+10e^-x=0. Then, by multiplying both sides of the equation by e^x, we can get an equation in the form of a quadratic equation: e^2x-7e^x+10=0. By setting y = e^x, 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 e^x back into the equation and solve for x using the ln laws: e^x = 2 or 5; therefore x = ln2 or ln5.