How do you solve ${log}_{a} (X + 4) - {log}_{a} (X - 7) = {log}_{a} (X - 8) - {log}_{a} (X - 9)$ $log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)$ ?

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How do you solve ${log}_{a} (X + 4) - {log}_{a} (X - 7) = {log}_{a} (X - 8) - {log}_{a} (X - 9)$ $log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)$ ?

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Answer 1 · 2015-07-24T19:52:12

$X = \frac{46}{5} = 9.2$ $X=46/5=9.2$

Explanation:

First, use the property that ${log}_{a} (\frac{C}{D}) = {log}_{a} (C) - {log}_{a} (D)$ $log_{a}(C/D)=log_{a}(C)-log_{a}(D)$ to rewrite the equation as

${log}_{a} (\frac{X + 4}{X - 7}) = {log}_{a} (\frac{X - 8}{X - 9})$ $log_{a}((X+4)/(X-7))=log_{a}((X-8)/(X-9))$

Next, use the fact that ${log}_{a} (z)$ $log_{a}(z)$ is a one-to-one function (when $a > 0$ $a>0$ and $a \neq 1$ $a!=1$ ) to say that the inputs are the same:

$\frac{X + 4}{X - 7} = \frac{X - 8}{X - 9}$ $(X+4)/(X-7)=(X-8)/(X-9)$

Now cross-multiply to get:

$(X + 4) (X - 9) = (X - 8) (X - 7)$ $(X+4)(X-9)=(X-8)(X-7)$

After using FOIL, this becomes

$X^{2} - 5 X - 36 = X^{2} - 15 X + 56$ $X^{2}-5X-36=X^2-15X+56$ .

Rearranging and canceling appropriately gives

$10 X = 92$ $10X=92$ so that $X = \frac{92}{10} = \frac{46}{5} = 9.2$ $X=92/10=46/5=9.2$

You should always check your answer in the original equation. When $X = 9.2$ $X=9.2$ , we have

$X + 4 = 13.2$ $X+4=13.2$ , $X - 7 = 2.2$ $X-7=2.2$ , $X - 8 = 1.2$ $X-8=1.2$ , and $X - 9 = 0.2$ $X-9=0.2$

You can use any value for $a > 0$ $a>0$ except $a = 1$ $a=1$ . For instance, if we use $a = 10$ $a=10$ , then:

${log}_{10} (13.2) \approx 1.120574$ $log_{10}(13.2)approx 1.120574$ , ${log}_{10} (2.2) \approx 0.342423$ $log_{10}(2.2)approx 0.342423$ , ${log}_{10} (1.2) \approx 0.079181$ $log_{10}(1.2)approx 0.079181$ , and ${log}_{10} (0.2) \approx - 0.698970$ $log_{10}(0.2)approx -0.698970$

and ${log}_{10} (13.2) - {log}_{10} (2.2) \approx 1.120574 - 0.342423 = 0.778151$ $log_{10}(13.2)-log_{10}(2.2)approx 1.120574-0.342423= 0.778151$ and ${log}_{10} (1.2) - {log}_{10} (0.2) \approx 0.079181 - (- 0.698970) = 0.778151$ $log_{10}(1.2)-log_{10}(0.2)approx 0.079181-(-0.698970)=0.778151$

Answer 2 · 2015-07-24T20:07:41

I found $x = 9.2$ $x=9.2$

Explanation:

You can use the fact that: ${log}_{a} x - {log}_{a} y = {log}_{a} (\frac{x}{y})$ $log_ax-log_ay=log_a(x/y)$ to get:
${log}_{a} (\frac{x + 4}{x - 7}) = {log}_{a} (\frac{x - 8}{x - 9})$ $log_a((x+4)/(x-7))=log_a((x-8)/(x-9))$
take the power of $a$ $a$ on bothe sides to get rid of the logs:
$cancel(a)^(cancel(log_a)((x+4)/(x-7)))=cancel(a)^(cancel(log_a)((x-8)/(x-9)))$
$(x+4)/(x-7)=(x-8)/(x-9)$
$(x+4)(x-9)=(x-8)(x-7)$
$cancel(x^2)-9x+4x-36=cancel(x^2)-7x-8x+56$
$10x=92$
$x=9.2$

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How do you solve ${log}_{a} (X + 4) - {log}_{a} (X - 7) = {log}_{a} (X - 8) - {log}_{a} (X - 9)$ $log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)$ ?

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How do you solve log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)loga(X+4)−loga(X−7)=loga(X−8)−loga(X−9)log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)?

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How do you solve ${log}_{a} (X + 4) - {log}_{a} (X - 7) = {log}_{a} (X - 8) - {log}_{a} (X - 9)$ $log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)$ ?