2 Logarithmic function problems

[Linty Fresh]

How would I go about solving the following?

Exact value of log2(6) * log6(4)

Also, I'm having trouble solving this one:

Solve for:
ln (x) + ln (x+2) = 4

I'm combining the terms and coming up with something like:
ln [x(x+2)]=4 and going to:

e^4=x(x+2) and then trying either multiplying the thing out and transfering e to get:

x^2+2x+3^4=0

or taking the natural logs like so:

ln e^4=ln[x(x+2)]
and simplifying to
(4)(ln e)=ln[x(x+2)]

But I'm stumped at this stage. Thanks so much for your help.

 

[tkhunny]

Exact value of log2(6) * log6(4)

You need this: log_b(a) = log_c(a)/log_c(b)
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[stapel]

By the way, e does not equal 3.

Eliz.

 

[Linty Fresh]

Oh, CRUD, that's how you do it. It did not occur to me to just write out the Change of Base Formula and that the log-base-6's would cancel themselves out, leaving me with log(4)/log(2). I forgot to mention that the first problem states not to use calculators, although I can use them in the second problem.

Thanks, tk

 

[Linty Fresh]

By the way, e does not equal 3.

Eliz.

Thanks for the correction, stapel. I meant to type:

x^2+2x+e^4=0

Any ideas? Thanks.

 

[tkhunny]

Isn't that MINUS e⁴?

It's just a quadratic equation. Don't let the 'e' scare you.

 

[soroban]

Hello, Linty Fresh!

There is an obscure theorem which busts the first problem.

. . . . . . . . . . . . . . 1
. . . log_b(N) .= .---------- . . ---> . . log_b(N) * log_N(b) . = . 1
. . . . . . . . . . . . log_N(b)

In baby-talk: If you switch the base and the argument, you get a reciprocal.

Exact value of log₂(6)*log₆(4)
.

. . Note that: . log₆(4) .= .log₆(2²) .= .2*log₆(2)

So we have: . log₂(6)*log₆(4) . = . log₂(6)*2*log₆(2)

. . . . . . . . . = . 2*[log₂(6)*log₆(2)] . = . 2[1] . = . 2

 

[rahian2k]

[deleted]

 

[Linty Fresh]

Thanks, tk and rahian2k.

I actually solved the first one this morning before work. Yeah, I somehow mistook e for a variable for a little while. Once it hit me over my morning coffee that it was a constant, I just plugged the whole thing into the quadratic formula. Thanks again for your help, and I'll be asking again soon!