Log

1) Solve the equation e^2x + 4e^(-2x) = 4

Ans: x = 0.347


2) Solve the equation:

5^x = 2^(x+3)


Ans: 2.27


Could anyone help me with this 2 questions? Thanks a lot. :)

e^2x+4e^-2x = 4
Multiply all by e^2x:
e^4x+ 4 = 4e^2x

Let y = e^2x

y² - 4y + 4 = 0

The solution is y = 2 => x = 0.5*ln(y) = 0.347...

Quote:

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Originally Posted by krtxmrtz

e^2x+4e^-2x = 4
Multiply all by e^2x:
e^4x+ 4 = 4e^2x

Let y = e^2x

y² - 4y + 4 = 0

The solution is y = 2 => x = 0.5*ln(y) = 0.347...

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Hi krtxmrtz,

Firstly, thanks for your help.

Secondly, may I know why you need to multiply all by e^2x ? Can't we just let y = e^2x ?

Also, if I multipy 4e^ (-2x) with e^2x , why the answer is 4? Because there is a negative (-2x) in 4e^(-2x).

I am confused about the above 2 points. Could you explain to me? :)

Thanks thanks. :D

You can let y = e^2x if you like. But you will then get the equation:

y + 4/y = 4

which, to solve, you will need to multiply through by y. i.e. it is the same thing.

When you multiply things with powers together, you must add the powers. Think of it this way, 2³ x 2 = 2³⁺¹ = 2⁴

So e^-2x x e^2x = e⁰ = 1


zaza

5^x = 2^(x+3) = 2^x * 2³ = 8*2^x

So you apply logarithms and get

x log(5) = log(8) + x log(2)

where log is the logarithm (base 10 or any other base)

and I assume you know how to deal with the rest.

Quote:

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Originally Posted by zaza

You can let y = e^2x if you like. But you will then get the equation:

y + 4/y = 4

which, to solve, you will need to multiply through by y. i.e. it is the same thing.

When you multiply things with powers together, you must add the powers. Think of it this way, 2³ x 2 = 2³⁺¹ = 2⁴

So e^-2x x e^2x = e⁰ = 1


zaza

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Now I understand. Thanks so much zaza. :D

Quote:

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Originally Posted by krtxmrtz

5^x = 2^(x+3) = 2^x * 2³ = 8*2^x

So you apply logarithms and get

x log(5) = log(8) + x log(2)

where log is the logarithm (base 10 or any other base)

and I assume you know how to deal with the rest.

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Hmm krtxmrtz, how do I get rid of the x in x log(2)? There are two (x)s in x log(2) & x log(5).

Quote:

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Originally Posted by Yunie

Hmm krtxmrtz, how do I get rid of the x in x log(2)? There are two (x)s in x log(2) & x log(5).

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x log(5) = log(8) + x log(2)
x log(5) - x log(2) = log(8)
x [log(5) - log(2)] = log (8)

x = log(8) / [log(5) - log(2)] = 2.269 (approx.)

Quote:

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Originally Posted by krtxmrtz

x log(5) = log(8) + x log(2)
x log(5) - x log(2) = log(8)
x [log(5) - log(2)] = log (8)

x = log(8) / [log(5) - log(2)] = 2.269 (approx.)

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Thanks so much krtxmrtx! ;) You are a great help. :D