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trasher [3.6K]
3 years ago
13

Solve the following equations. log2(x^2 − 16) − log^2(x − 4) = 1

Mathematics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

x=\frac{4*(2+e)}{e-2}

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

log(\frac{1}{x} )=-log(x)

log(x*y)=log(x)+log(y)

Therefore:

log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1

Now, cancel logarithms by taking exp of both sides:

e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e

Multiply both sides by (x-4)^{2} and using distributive propierty:

2x^{2} -32=16e-8ex+ex^{2}

Substract 16e-8ex+ex^{2} from both sides and factoring:

-(x-4)*(-8-4e-2x+ex)=0

Multiply both sides by -1:

(x-4)*(-8-4e-2x+ex)=0

Split into two equations:

x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0

Solving for x-4=0

Add 4 to both sides:

x=4

Solving for -8-4e-2x+ex=0

Collect in terms of x and add 4e+8 to both sides:

x(e-2)=4e+8

Divide both sides by e-2:

x=\frac{4*(2+e)}{e-2}

The solutions are:

x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}

If we evaluate x=4 in the original equation:

log(0)-log(0)=1

This is an absurd because log (x) is undefined for x\leq 0

If we evaluate x=\frac{4*(2+e)}{e-2} in the original equation:

log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1

Which is correct, therefore the solution is:

x=\frac{4*(2+e)}{e-2}

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