# Exponent Question ?

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Would someone be able to solve this equation for me: 3e^-10x = 6

i just want to see if my solution is right... thanks.

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And what would that solution be?

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The route to a solution can have many paths . Some take one step where others take three . Some explain one step where others explain three . When you put a calculated value for x into the equation do you get an answer of 6 ?

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first thing to do would be simplify it and make it clear to the board where the exponent ends. the next step is fairly easy - but fswd is correct; show us what you got already

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It is possible to make a mistake and another mistake which then corrects it and you don't know of either . When you manipulate the original equation you will get different equations . Put the final value of x that you calculated at the very end into each of these equations to see if the left hand side equals the right hand side all the way from start to finish .

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ok, i got: x = ln(2/10)...

ok, i got: x = ln(2/10)...

but my answer doesn't give me 6 when subbed back into the equation.

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x is not equal to ln(2/10)

I think you have dropped a negative sign somewhere along the way .

You may also be getting the natural log of 10 % of what you should be getting the natural log of .

Edited by Hal.

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ok, i got: x = ln(2/10)...

but my answer doesn't give me 6 when subbed back into the equation.

i got: x = ln(2/10)...

Show your entire equation. There's no use in speculating about an equation with "..." in it.

Edited by ewmon

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Finally , I think you brought the 10 into the denominator of what you want to find the natural log of , when it should be a 1/10 multiplication factor in a product with your natural log .

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1123581321 - you are not really gonna learn unless you put down your method. the comments from forum members are all correct as far as I can see, but they cannot really help till you are more explicit. maths just doesn't work in leaps - it works in small repeatable and easily understood steps; I would do your equation in 5 steps, even though I know what the answer would be in 1 step.

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The equation of the form $e^{\beta x} = \alpha$ have a real solution of $x = \frac{log(\alpha)}{\beta}$,

Edited by khaled

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