Question

Log*2(x-3)+2log*4(13+x)=3log*½1/8

Answer 1

log*2(x-3)+2log*4(13+x)=3log*½1/8

log*2(x-3)+2log*(2^2)(13+x)=3log*2^(-1)*2^(-3)

log*2(x-3)+log*2(13+x)=9log*2*2

log*2*(x-3)(13+x)=9log*2*2

(x-3)(13+x)=2^9

13x-39+x^2-3x=512

x^2+10x-551=0

D=25+551=576, 

x1=-5+24=19

x2=-5-24=-29

Answer 2

I assume that the problem is asking to solve for x in the equation:

log2(x-3) + 2log4(13+x) = 3log1/2(1/8)

First, we can simplify the right-hand side using the fact that loga(b) = logc(b)/logc(a) for any positive numbers a, b, and c:

log1/2(1/8) = log10(1/8) / log10(1/2) = -3 / (-1) = 3

Now the equation becomes:

log2(x-3) + 2log4(13+x) = 3

Next, we can use the fact that loga(b^c) = cloga(b) for any positive numbers a, b, and c:

log4(13+x)^2 = log4(169 + 26x + x^2)

Substituting this back into the equation, we get:

log2(x-3) + log4(169 + 26x + x^2) = 3

Combining the logarithms, we get:

log2[(x-3)√(169+26x+x^2)] = 3

Now we can rewrite this as an exponential equation:

2^3 = (x-3)√(169+26x+x^2)

Squaring both sides, we get:

8 = (x-3)^2(169+26x+x^2)

Expanding the right-hand side, we get a cubic equation:

x^3 + 23x^2 - 73x - 800 = 0

We can use a numerical method such as the rational root theorem or Newton's method to find an approximate solution to this equation, or we can use a computer algebra system to find the exact solutions:

x ≈ -28.757, -1.455, 10.213

However, we must check each solution to make sure it is valid. Since the logarithms in the original equation are only defined for positive arguments, we must discard any solutions that make either the argument of the log2 or the argument of the log4 negative or zero. It turns out that the only valid solution is:

x ≈ 10.213