Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$
$begingroup$
Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$
I equated their exponents,
That gave me $log_5 x = frac{8}{3}$
But the answer given in my book is $1$.
Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.
When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?
Any help would be appreciated.
functions logarithms
$endgroup$
add a comment |
$begingroup$
Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$
I equated their exponents,
That gave me $log_5 x = frac{8}{3}$
But the answer given in my book is $1$.
Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.
When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?
Any help would be appreciated.
functions logarithms
$endgroup$
$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36
add a comment |
$begingroup$
Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$
I equated their exponents,
That gave me $log_5 x = frac{8}{3}$
But the answer given in my book is $1$.
Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.
When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?
Any help would be appreciated.
functions logarithms
$endgroup$
Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$
I equated their exponents,
That gave me $log_5 x = frac{8}{3}$
But the answer given in my book is $1$.
Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.
When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?
Any help would be appreciated.
functions logarithms
functions logarithms
edited Dec 8 '18 at 11:48
N. F. Taussig
44k93356
44k93356
asked Dec 6 '18 at 16:05
Piano LandPiano Land
379115
379115
$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36
add a comment |
$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36
$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36
$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.
On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$
$endgroup$
add a comment |
$begingroup$
I suppose you were using
$$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
but it is only true when $x>0$ and $xne 1$.
Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.
$endgroup$
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
add a comment |
$begingroup$
We need
$$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$
that is
$$x=5^frac83$$
the other solution $x=1$ is obtained by inspection from the original equation.
$endgroup$
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
add a comment |
$begingroup$
If the book claims, that $1$ is the only real solution, it is wrong. Your value
$$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.
$endgroup$
add a comment |
$begingroup$
What you did (probably) was the following:
$$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$
By doing so, you removed the possibility of $x = 1$.
As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.
Your other solution is valid. Let $x = 5^{frac{8}{3}}$.
$$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$
On both sides, you get
$$big(5^{frac{8}{3}}big)^{-4}$$
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.
On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$
$endgroup$
add a comment |
$begingroup$
You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.
On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$
$endgroup$
add a comment |
$begingroup$
You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.
On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$
$endgroup$
You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.
On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$
answered Dec 6 '18 at 16:49
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
I suppose you were using
$$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
but it is only true when $x>0$ and $xne 1$.
Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.
$endgroup$
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
add a comment |
$begingroup$
I suppose you were using
$$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
but it is only true when $x>0$ and $xne 1$.
Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.
$endgroup$
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
add a comment |
$begingroup$
I suppose you were using
$$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
but it is only true when $x>0$ and $xne 1$.
Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.
$endgroup$
I suppose you were using
$$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
but it is only true when $x>0$ and $xne 1$.
Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.
answered Dec 6 '18 at 16:11
Eclipse SunEclipse Sun
6,9941437
6,9941437
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
add a comment |
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
$endgroup$
– Piano Land
Dec 6 '18 at 16:25
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:28
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
$begingroup$
Perhaps just because $x=5^{8/3}$ is too large.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 16:38
add a comment |
$begingroup$
We need
$$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$
that is
$$x=5^frac83$$
the other solution $x=1$ is obtained by inspection from the original equation.
$endgroup$
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
add a comment |
$begingroup$
We need
$$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$
that is
$$x=5^frac83$$
the other solution $x=1$ is obtained by inspection from the original equation.
$endgroup$
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
add a comment |
$begingroup$
We need
$$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$
that is
$$x=5^frac83$$
the other solution $x=1$ is obtained by inspection from the original equation.
$endgroup$
We need
$$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$
that is
$$x=5^frac83$$
the other solution $x=1$ is obtained by inspection from the original equation.
edited Dec 6 '18 at 16:27
answered Dec 6 '18 at 16:24
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
add a comment |
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
@KM101 Opssss...thanks I fix!
$endgroup$
– gimusi
Dec 6 '18 at 16:26
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
$begingroup$
No problem! :-)
$endgroup$
– KM101
Dec 6 '18 at 16:36
add a comment |
$begingroup$
If the book claims, that $1$ is the only real solution, it is wrong. Your value
$$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.
$endgroup$
add a comment |
$begingroup$
If the book claims, that $1$ is the only real solution, it is wrong. Your value
$$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.
$endgroup$
add a comment |
$begingroup$
If the book claims, that $1$ is the only real solution, it is wrong. Your value
$$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.
$endgroup$
If the book claims, that $1$ is the only real solution, it is wrong. Your value
$$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.
edited Dec 6 '18 at 16:28
answered Dec 6 '18 at 16:22
gammatestergammatester
16.7k21632
16.7k21632
add a comment |
add a comment |
$begingroup$
What you did (probably) was the following:
$$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$
By doing so, you removed the possibility of $x = 1$.
As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.
Your other solution is valid. Let $x = 5^{frac{8}{3}}$.
$$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$
On both sides, you get
$$big(5^{frac{8}{3}}big)^{-4}$$
$endgroup$
add a comment |
$begingroup$
What you did (probably) was the following:
$$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$
By doing so, you removed the possibility of $x = 1$.
As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.
Your other solution is valid. Let $x = 5^{frac{8}{3}}$.
$$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$
On both sides, you get
$$big(5^{frac{8}{3}}big)^{-4}$$
$endgroup$
add a comment |
$begingroup$
What you did (probably) was the following:
$$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$
By doing so, you removed the possibility of $x = 1$.
As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.
Your other solution is valid. Let $x = 5^{frac{8}{3}}$.
$$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$
On both sides, you get
$$big(5^{frac{8}{3}}big)^{-4}$$
$endgroup$
What you did (probably) was the following:
$$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$
By doing so, you removed the possibility of $x = 1$.
As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.
Your other solution is valid. Let $x = 5^{frac{8}{3}}$.
$$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$
On both sides, you get
$$big(5^{frac{8}{3}}big)^{-4}$$
edited Dec 6 '18 at 16:36
answered Dec 6 '18 at 16:22
KM101KM101
5,9281524
5,9281524
add a comment |
add a comment |
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$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36