log again: domain of f(x) = log_2(100-x^2); solving log eqns

[brittany6990]

f(x)= log[sub:5vq8n4i7]2[/sub:5vq8n4i7] (100- x[sup:5vq8n4i7]2[/sup:5vq8n4i7])

would the domain of this function be x < 10 or x < 100?

and if anyone knows how to do:

log[sub:5vq8n4i7]3[/sub:5vq8n4i7] x+ log [sub:5vq8n4i7]3[/sub:5vq8n4i7](x-24) = 4

and

5[sup:5vq8n4i7]x+7[/sup:5vq8n4i7]=7


I don't know how to go about solving these. help! Thank you!

 

[Deleted member 4993]

Re: log again

f(x)= log[sub:3axfsm2v]2[/sub:3axfsm2v] (100- x[sup:3axfsm2v]2[/sup:3axfsm2v])

would the domain of this function be x < 10 or x < 100?

The domain of a log function is restricted to the positive values of the "argument".

Then we have:

100 - x[sup:3axfsm2v]2[/sup:3axfsm2v] > 0

solve above....
and if anyone knows how to do:

log[sub:3axfsm2v]3[/sub:3axfsm2v] x+ log [sub:3axfsm2v]3[/sub:3axfsm2v](x-24) = 4

log[sub:3axfsm2v]a[/sub:3axfsm2v] x+ log [sub:3axfsm2v]a[/sub:3axfsm2v](x+b) = c

log[sub:3axfsm2v]a[/sub:3axfsm2v] {x(x+b)} = c

x(x+b) = a[sup:3axfsm2v]c[/sup:3axfsm2v]

x[sup:3axfsm2v]2[/sup:3axfsm2v] + bx - a[sup:3axfsm2v]c[/sup:3axfsm2v] = 0

The above is a quadratic equation - solve for 'x'....
and

5[sup:3axfsm2v]x+7[/sup:3axfsm2v]=7

a[sup:3axfsm2v]x+b[/sup:3axfsm2v] = c ..... Take "log" of both sides

log{a[sup:3axfsm2v]x+b[/sup:3axfsm2v]} = log(c)

(x+b) * log(a) = log(c)

solve for 'x' from above...

I don't know how to go about solving these. help! Thank you!