Can someone please help me solve these logarithmic equations?

Question

I keep trying/getting them wrong - can someone sort of show me how to get through them? I'd really appreciate it =]

log5(x+20) + log5(x) = 3

and

log2(3x+20) - 2log2(x) = 3

(The numbers right after the logs are bases)

Answer 1

Let's try it without using the usual log relation laws (which are new to people at this stage and thus easily forgotten) and instead use the exponential relations which are much more well known by this stage.

For the first one take 5^LHS(i.e. left hand side) and 5^RHS(i.e. right hand side)

5^(log5(x+20) + log5(x)) = 5^3 = 125

Now we know a^(b+c)=(a^b)*(a^c) so:

(5^(log5(x+20)))*(5^(log5(x))) = 125

^ and log are inverse operators i.e. x=a^loga(x)=loga(a^x) thus:

(x+20)*(x) = 125

x^2 +20x -125 = 0

we thus have a quadratic equation. This factorises as:

(x-5)(x+25) = 0

which has solutions x = 5 or x = -25

For the second I'll do the working out a bit quicker as it parallels the first (except for the -2 outside the second term which transforms as: -2log2(x) = log2(x^(-2)) if you don't remember this relation it can be done exponentially as -2log2(x) = -log2(x) -log2(x) and the minus term will turn to divide bys when expanding the exponent)

(3x+20)/(x^2) = 2^3 = 8

8x^2 - 3x - 20 = 0 (*Cheryl B made her mistake here* and as a result there are real solutions*)

this doesn't factorise as nicely so we use the quadratic equation:

x = -b/2a +/- (sqrt((b^2) - 4ac))/2a

   = (3 +/- sqrt(649)))/16

Answer 2

log5(x+20) + log5(x) can be combined into

log5(x+20)*x

log5(x+20)*x=3

(x+20)*x=5^3

x^2+20x=125

x^2+20x-125=0

(x+25)(x-5)=0

x=-25 or x=5

Throw out the negative answer because the log of a negative number is undefined.  x=5

log2(3x+20)-2log2(x)=3

log2(3x+20)-log2(x^2)=3

log2[(3x+20)/x^2]=3

(3x+20)/x^2=2^3

3x+20=8x^2

0=8x^2-3x+20

Here, b^2-4ac=9-4*8*20, which is negative, so there are no real solutions.

_/

Answer 3

first one, multiply x and x plus 20

so log5(x squared and 20 x) = 3

simplify

5^3 = x squared and 20x

125 = x^2 + 20x

put into a quadratic equation, factor and solve

second one's a bit more complicated

get rid of the 2 first

log2(3x+20) - log2(x^2) = 3

then divide

log 2 (3x+20/x^2) = 3

so 8 = 3x+20 / x^2

8x^2 = 3x + 20

8x^2 - 3x - 20 =0

factor and solve

Answer 4

Ah, you are missing the forest because of the trees.  So look at the trees; what does an sum of two log terms mean?

I hope there is an "AHA" to that one, for that is the key to the problem.   Then, you take antilogs of both sides which (in case you didn't "AHA") is., for the first problem,...........................

  (x+20)(x)=125.   WAS IT REALLY THAT EASY?  Well, there is still a quadratic to play with (negative answers are a no,no).