Log

[Yunie]

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.

[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...

[Yunie]

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...

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.

[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

[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.

[Yunie]

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

Now I understand. Thanks so much zaza.

[Yunie]

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.

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).

[krtxmrtz]

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).

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.)

[Yunie]

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.)

Thanks so much krtxmrtx! You are a great help.