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# Help with Logarithms Questions! watch

1. I am stuck on these logs questions, if anyone can offer any assistance I will really appreciate it and give positive rep, thanks

1) Find the root of the equation 10^2-2x = 2 x 10^-x giving your answer exactly in terms of logarithms.
2) Express log (2 root 10) - 1/3 log 0.8 - log(10/3) in the form c + log d where c and d are rational numbers and the logarithms are to base 10.
3) Solve the equation e^0.5x+1 = 8, giving your answer in terms of log[to base e] 2.

Right so for question 1, I have done:

(2 - 2x)log 10 = (-x)log 20
2-2x = -x log20
2 = -xlog20 + 2x

I don't know how to proceed from here

For question 2 I got up to:

log (2 root 10) - log (0.8^1/3) - log (10/3)

Don't know what to do next.

And for question 3 I have done:

log[to base e] 8 = 0.5x+1
8log[to base e] = 0.5x+1

Don't know what to do now

Any thoughts??? Thanks!
2. 1) log of 2*10^-x is log2 + log(10^-x) = log2 - xlog10

3) e^(0.5x) +1 =8
e^(0.5x) = 7
Now log base e both sides
3. (Original post by vc94)
1) log of 2*10^-x is log2 + log(10^-x) = log2 - xlog10

3) e^(0.5x) +1 =8
e^(0.5x) = 7
Now log base e both sides
How does that help?

And for question 3, you cannot take away a power....It's e^(0.5x+1) = 8.
4. 3) Ok, so you meant e^(0.5x +1) =8.
Log base e both side gives 0.5x+1=ln8 where ln means log base e.
Make x the subject noting that 8=2^3
5. 1) (2 - 2x)log 10 = log2 - xlog10
Multiply the brackets, rearrange to make x the subject.
6. (Original post by vc94)
1) (2 - 2x)log 10 = log2 - xlog10
Multiply the brackets, rearrange to make x the subject.
So then you get:

2log10 - 2xlog10 = log2 - xlog10
2log10 - log2 = xlog10 + 2xlog10
2log10 - log2 = x (log10 + 2log10)
2 - log2 = x(1+2)
x = 2 - log 2 /3

?
7. (Original post by Nator)
So then you get:

2log10 - 2xlog10 = log2 - xlog10
2log10 - log2 = xlog10 + 2xlog10
2log10 - log2 = x (log10 + 2log10)
2 - log2 = x(1+2)
x = 2 - log 2 /3

?
Yes, working with log base 10.

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Updated: March 19, 2011
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