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Firstly set 2sinhx+3coshx=5

Now using the exponential definitions of sinhx and coshx rewrite the equation to give:

2(1/2(e^x-e^-x))+3(1/2(e^x+e^-x))=5

Simplify the equation by expanding out the brackets, multipling by 2 to eliminate fractions and collecting like terms together, as so:

e^x-e^-x+3/2e^x+3/2e^-x=5

2e^x-2e^-x+3e^x+3e^-x=10

5e^x+e^-x=10

5e^x+e^-x-10=0

e^-x is equivalent to 1/e^x therefore multiply through by e^x to get a quadratic equation in e^x

5e^2x-10e^x+1=0

Now using the quadratic equation (where a=5, b=-10 and c=1) solve for e^x

I will indicate 'plus or minus' by +/- (not to be confused with plus, divide, minus)

e^x=(-(-10)+/-√(-10)^2-4(5)(1))/2(5)

e^x=(10+/-√80)/10

e^x=1+/-(2√5)/5

To solve for x you must take the natural logarithm of both sides as (ln^e=1) so

x=ln(1+(2√5)/5

or

x=ln(1-(2√5)/5)

11 months ago

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