Given ${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x−y)=1$ then $y = a x^{n} + x + b$ $y=ax^n+x+b$ , what is $a$ $a$ , $b$ $b$ , $n$ $n$ ?

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Given ${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x−y)=1$ then $y = a x^{n} + x + b$ $y=ax^n+x+b$ , what is $a$ $a$ , $b$ $b$ , $n$ $n$ ?

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Answer 1 · 2018-06-07T07:48:43

$a = 5$ $a=5$
$n = 2$ $n=2$
$b = - 3$ $b=-3$

Explanation:

Using the rule of logs, ${log}_{a} (a) = 1$ $log_a(a)=1$ , we know that $5 x^{2} + x - y = 3$ $5x^2+x-y=3$

$y = 5 x^{2} + x - 3$ $y=5x^2+x-3$

$a = 5$ $a=5$
$n = 2$ $n=2$
$b = - 3$ $b=-3$

Answer 2 · 2018-06-07T07:49:12

See explanatio.

Explanation:

${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x-y)=1$

${log}_{3} (5 x^{2} + x - y) = {log}_{3} 3$ $log_3(5x^2+x-y)=log_3 3$

$5 x^{2} + x - y = 3$ $5x^2+x-y=3$

$y = 5 x^{2} + x - 3$ $y=5x^2+x-3$

So we can write that:

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Given ${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x−y)=1$ then $y = a x^{n} + x + b$ $y=ax^n+x+b$ , what is $a$ $a$ , $b$ $b$ , $n$ $n$ ?

2 Answers

Explanation:

Explanation:

${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x-y)=1$

${log}_{3} (5 x^{2} + x - y) = {log}_{3} 3$ $log_3(5x^2+x-y)=log_3 3$

$5 x^{2} + x - y = 3$ $5x^2+x-y=3$

$y = 5 x^{2} + x - 3$ $y=5x^2+x-3$

$a = 5, b = - 3, n = 2$ $a=5,b=-3, n=2$

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Given log3(5x2+x−y)=1log_3(5x^2+x−y)=1 then y=axn+x+by=ax^n+x+b, what is aa, bb, nn?

2 Answers

Explanation:

Explanation:

log3(5x2+x−y)=1log_3(5x^2+x-y)=1

log3(5x2+x−y)=log33log_3(5x^2+x-y)=log_3 3

5x2+x−y=35x^2+x-y=3

y=5x2+x−3y=5x^2+x-3

a=5,b=−3,n=2a=5,b=-3, n=2

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Impact of this question

Given ${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x−y)=1$ then $y = a x^{n} + x + b$ $y=ax^n+x+b$ , what is $a$ $a$ , $b$ $b$ , $n$ $n$ ?

${log}_{3} (5 x^{2} + x - y) = 1$ $log_3(5x^2+x-y)=1$

${log}_{3} (5 x^{2} + x - y) = {log}_{3} 3$ $log_3(5x^2+x-y)=log_3 3$

$5 x^{2} + x - y = 3$ $5x^2+x-y=3$

$y = 5 x^{2} + x - 3$ $y=5x^2+x-3$

$a = 5, b = - 3, n = 2$ $a=5,b=-3, n=2$