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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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#1 h Apr 16, 2025, 3:09 AM

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4b+13/ -4b+15 = 1/6

anyone can help on the steps to do this ? thank u

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That_One_Genius.

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That_One_Genius.

#2 h Apr 16, 2025, 3:42 AM

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Since you know that the ratio of both sides is 1:6, you can multiply both sides by 6 to get 24b+78/-4b+15 = 1 or 24b+78=-4b+15. Then it's just a matter of simplifying. 24b+78+(-15)=-4b+15+(-15) -> 24b+63+(-63)=-4b+(-63) -> 24b = -4b-63 -> 24b + (4b)=-4b+(4b)-63 -> 28b = -63 -> b = -63/28.

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#3 h Apr 16, 2025, 4:06 AM

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thank u, but the original question is
log₅ (4b+13) - log₅ (-4b+ 15) = log₅ 1/b
sorry my mistyping the equal above. it is 1/b

the answer is b = 3/4

but I don't know how question comes to this answer

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#4 h Apr 16, 2025, 4:07 AM

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I have returned after one year.

$\dfrac{4b+13}{-4b+15} = \dfrac{1}{6}$ $6(4b+13) = -4b+15$ $24b+78 = -4b+15$ $28b=-63$ $b = -\dfrac{9}{4}$
edit: I did not see the correction of the problem. Refer to my later solution.

This post has been edited 1 time. Last edited by OofPirate, Apr 16, 2025, 4:15 AM

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maxamc

#5 h Apr 16, 2025, 4:11 AM • 1 Y

Y by Miranda2829

Miranda2829 wrote:

thank u, but the original question is
log₅ (4b+13) - log₅ (-4b+ 15) = log₅ 1/b
sorry my mistyping the equal above. it is 1/b

the answer is b = 3/4

but I don't know how question comes to this answer

$\log_5(4b+13)-\log_5(-4b+ 15)=\log_5\frac{1}{b} \implies \frac{4b+13}{-4b+15}=\frac{1}{b} \implies 4b^2+13b=-4b+15 \implies 4b^2+17b-15=0.$

Solving gives $b=\frac{3}{4}$ , and $b=-5$ , the latter is a degenerate solution. We are done. $\blacksquare$

This post has been edited 2 times. Last edited by maxamc, Apr 16, 2025, 4:14 AM

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#6 h Apr 16, 2025, 4:14 AM • 1 Y

Y by Miranda2829

Miranda2829 wrote:

thank u, but the original question is
log₅ (4b+13) - log₅ (-4b+ 15) = log₅ 1/b
sorry my mistyping the equal above. it is 1/b

the answer is b = 3/4

but I don't know how question comes to this answer

If it's $\dfrac{1}{b}$ , then $\log_5 (4b+13) - \log_5 ( -4b+15) = \log_5 (1/b) = \log_5 (b^{-1}) = -log_5 (b)$
Thus $\log_5 (\dfrac{4b+13}{-4b+15}) = -log_5 (b)$ , and $\log_5 (\dfrac{b(4b+13)}{-4b+15}) = 0$

Thus, $\dfrac{b(4b+13)}{-4b+15}=1$ , and $4b^2+13b=-4b+15$ , and $4b^2+17b-15 = 0$ . This becomes $(4x-3)(x+5)$ . Answer is b=3/4 and b=-5. However, it is impossible to have logarithm of negative number, thus b=-5 is undefined. Answer is $b=\boxed{\dfrac{3}{4}}$

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