Question 692608
{{{-2*e^(9.4x-5)+6=-68.9}}}
First we isolate the base and its exponent. Subtracting 6 we get:
{{{-2*e^(9.4x-5)=-74.9}}}
Dividing by -2:
{{{e^(9.4x-5)=37.45}}}<br>
Now we use logarithms. Logarithms of any base may be used. But there are advantages to choosing certain bases:<ul><li>Choosing a base of logarithm that matches the base of the exponent will result in the simplest possible expression for the solution.</li><li>Choosing a base of logarithm that your calculator "knows" (base 10, "log", or base e, "ln") will result in an expression which will be easy to convert to a decimal approximation.</li></ul>In this equation the base of the exponent is a base our calculator "knows" so we can get both benefits by choosing base e logarithms:
{{{ln(e^(9.4x-5))=ln(37.45)}}}
On the left side we can use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, which allows us to move the exponent of the argument out in front. (It is this very property that is the reason we use logarithms on equations like this. It lets use move the exponent, where the variable is, out in front where we can "get at" the variable with "regular" algebra.) Using this property we get:
{{{(9.4x-5)ln(e)=ln(37.45)}}}
By definition ln(e) = 1. (This is why matching the base of the logarithm to the base of the exponent results in the simplest expression.)
{{{9.4x-5=ln(37.45)}}}
Now we can solve for x. Adding 5:
{{{9.4x=ln(37.45)+5}}}
Dividing by 9.4:
{{{x=(ln(37.45)+5)/9.4}}}
This is an exact expression for the solution to your equation. If you want/need a decimal approximation get out your calculator and use it on the right side.