Solve following 4e^{1+3 x }- 9e^{5-2 x }= 0 .

**Solution**

Here the first step is to get one exponential on every side & then we'll divide both sides by one of them (that doesn't matter for the most part) therefore we'll have a quotient of two exponentials. Then the quotient can be simplified & finally we'll get both coefficients on the other side.

Doing all of this gives,

4e^{1+3 x }= 9e^{5-2 x}

e^{1+3 x} / e^{5-2 x} = 9/4

e1+3 x-(5-2 x) = 9/4

e^{5 x-4 }= 9/4

Note as well that while we said that it doesn't actually matter which exponential we divide out by doing it the way we did now we'll ignore a negative coefficient on the x. Not major issue, however those minus signs on coefficients are actually easy to lose on occasion.

Now this is in a form that we can deal with so here's the rest of the solution.

e^{5 x-4 }= 9/4

5x - 4 = ln ( 9/4 )

5x = 4 + ln ( 9/4 )

x = 1/5 (4 ( ln ( 9/4 )) = 0.9621860432

This equation contain single solution of x = 0.9622 .

Now let's see some equations which involve logarithms. The main property that we'll be using to solve these kinds of equations is,

b^{logbx} = x

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