Can I write $log_{2^3*5^5} x = frac{1}{3*5} log_{2*5} x$? [closed]
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Since there is a property where we can take take the power of base into division
algebra-precalculus
edited Nov 18 at 6:53
Toby Mak
3,32811128
asked Nov 18 at 6:49
Rohan
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closed as unclear what you're asking by Henno Brandsma, user302797, amWhy, user10354138, tatan Nov 24 at 19:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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down vote
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Since there is a property where we can take take the power of base into division
algebra-precalculus
edited Nov 18 at 6:53
Toby Mak
3,32811128
asked Nov 18 at 6:49
Rohan
1
closed as unclear what you're asking by Henno Brandsma, user302797, amWhy, user10354138, tatan Nov 24 at 19:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
I cannot read the equation. Is it $2^3*5^5(x)=frac{1}{3*5}text{log}_{(2*5)}(x)$?
– Joey Kilpatrick
Nov 18 at 6:52
@Rohan This is what MathJax looks like. If this is not what you wanted, you can change it yourself.
– Toby Mak
Nov 18 at 6:54
it's correct now
– Rohan
Nov 18 at 7:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Since there is a property where we can take take the power of base into division
algebra-precalculus
edited Nov 18 at 6:53
Toby Mak
3,32811128
asked Nov 18 at 6:49
Rohan
1
Since there is a property where we can take take the power of base into division
algebra-precalculus
algebra-precalculus
edited Nov 18 at 6:53
Toby Mak
3,32811128
asked Nov 18 at 6:49
Rohan
1
edited Nov 18 at 6:53
Toby Mak
3,32811128
asked Nov 18 at 6:49
Rohan
1
edited Nov 18 at 6:53
Toby Mak
3,32811128
edited Nov 18 at 6:53
Toby Mak
3,32811128
edited Nov 18 at 6:53
Toby Mak
3,32811128
3,32811128
asked Nov 18 at 6:49
Rohan
1
asked Nov 18 at 6:49
Rohan
1
asked Nov 18 at 6:49
Rohan
1
1
closed as unclear what you're asking by Henno Brandsma, user302797, amWhy, user10354138, tatan Nov 24 at 19:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Henno Brandsma, user302797, amWhy, user10354138, tatan Nov 24 at 19:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
I cannot read the equation. Is it $2^3*5^5(x)=frac{1}{3*5}text{log}_{(2*5)}(x)$?
– Joey Kilpatrick
Nov 18 at 6:52
@Rohan This is what MathJax looks like. If this is not what you wanted, you can change it yourself.
– Toby Mak
Nov 18 at 6:54
it's correct now
– Rohan
Nov 18 at 7:02
add a comment |
I cannot read the equation. Is it $2^3*5^5(x)=frac{1}{3*5}text{log}_{(2*5)}(x)$?
– Joey Kilpatrick
Nov 18 at 6:52
@Rohan This is what MathJax looks like. If this is not what you wanted, you can change it yourself.
– Toby Mak
Nov 18 at 6:54
it's correct now
– Rohan
Nov 18 at 7:02
I cannot read the equation. Is it $2^3*5^5(x)=frac{1}{3*5}text{log}_{(2*5)}(x)$?
– Joey Kilpatrick
Nov 18 at 6:52
I cannot read the equation. Is it $2^3*5^5(x)=frac{1}{3*5}text{log}_{(2*5)}(x)$?
– Joey Kilpatrick
Nov 18 at 6:52
@Rohan This is what MathJax looks like. If this is not what you wanted, you can change it yourself.
– Toby Mak
Nov 18 at 6:54
@Rohan This is what MathJax looks like. If this is not what you wanted, you can change it yourself.
– Toby Mak
Nov 18 at 6:54
it's correct now
– Rohan
Nov 18 at 7:02
it's correct now
– Rohan
Nov 18 at 7:02
add a comment |
2 Answers
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The answer is no.
Let $x=10$ and the $RHS$ will be $frac {1}{15}$
The $LHS$ then implies $$2^3 times 5^5 =10^{15}$$ which is not true.
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
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By the change of base formula:
$$log_{color{red}{b}} (x) = frac{log_color{blue}{d} (x)}{log_color{blue}{d} (color{red}{b})}$$
$$log_{color{red}{2 cdot 5}} (x) = frac{log_color{blue}{2^3 cdot 5^5} (x)}{log_color{blue}{2^3 cdot 5^5} (color{red}{2 cdot 5})}$$
$$log_{2^3 cdot 5^5} (x) = frac{1}{log_{2^3 cdot 5^5} ({2 cdot 5})} cdot log_{2 cdot 5} (x)$$
but $log_{2^3 cdot 5^5} ({2 cdot 5}) ne 15$.
answered Nov 18 at 7:07
Toby Mak
3,32811128
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The answer is no.
Let $x=10$ and the $RHS$ will be $frac {1}{15}$
The $LHS$ then implies $$2^3 times 5^5 =10^{15}$$ which is not true.
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
1
down vote
The answer is no.
Let $x=10$ and the $RHS$ will be $frac {1}{15}$
The $LHS$ then implies $$2^3 times 5^5 =10^{15}$$ which is not true.
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
1
down vote
up vote
1
down vote
The answer is no.
Let $x=10$ and the $RHS$ will be $frac {1}{15}$
The $LHS$ then implies $$2^3 times 5^5 =10^{15}$$ which is not true.
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
The answer is no.
Let $x=10$ and the $RHS$ will be $frac {1}{15}$
The $LHS$ then implies $$2^3 times 5^5 =10^{15}$$ which is not true.
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 7:37
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
up vote
0
down vote
By the change of base formula:
$$log_{color{red}{b}} (x) = frac{log_color{blue}{d} (x)}{log_color{blue}{d} (color{red}{b})}$$
$$log_{color{red}{2 cdot 5}} (x) = frac{log_color{blue}{2^3 cdot 5^5} (x)}{log_color{blue}{2^3 cdot 5^5} (color{red}{2 cdot 5})}$$
$$log_{2^3 cdot 5^5} (x) = frac{1}{log_{2^3 cdot 5^5} ({2 cdot 5})} cdot log_{2 cdot 5} (x)$$
but $log_{2^3 cdot 5^5} ({2 cdot 5}) ne 15$.
answered Nov 18 at 7:07
Toby Mak
3,32811128
add a comment |
up vote
0
down vote
By the change of base formula:
$$log_{color{red}{b}} (x) = frac{log_color{blue}{d} (x)}{log_color{blue}{d} (color{red}{b})}$$
$$log_{color{red}{2 cdot 5}} (x) = frac{log_color{blue}{2^3 cdot 5^5} (x)}{log_color{blue}{2^3 cdot 5^5} (color{red}{2 cdot 5})}$$
$$log_{2^3 cdot 5^5} (x) = frac{1}{log_{2^3 cdot 5^5} ({2 cdot 5})} cdot log_{2 cdot 5} (x)$$
but $log_{2^3 cdot 5^5} ({2 cdot 5}) ne 15$.
answered Nov 18 at 7:07
Toby Mak
3,32811128
add a comment |
up vote
0
down vote
up vote
0
down vote
By the change of base formula:
$$log_{color{red}{b}} (x) = frac{log_color{blue}{d} (x)}{log_color{blue}{d} (color{red}{b})}$$
$$log_{color{red}{2 cdot 5}} (x) = frac{log_color{blue}{2^3 cdot 5^5} (x)}{log_color{blue}{2^3 cdot 5^5} (color{red}{2 cdot 5})}$$
$$log_{2^3 cdot 5^5} (x) = frac{1}{log_{2^3 cdot 5^5} ({2 cdot 5})} cdot log_{2 cdot 5} (x)$$
but $log_{2^3 cdot 5^5} ({2 cdot 5}) ne 15$.
answered Nov 18 at 7:07
Toby Mak
3,32811128
By the change of base formula:
$$log_{color{red}{b}} (x) = frac{log_color{blue}{d} (x)}{log_color{blue}{d} (color{red}{b})}$$
$$log_{color{red}{2 cdot 5}} (x) = frac{log_color{blue}{2^3 cdot 5^5} (x)}{log_color{blue}{2^3 cdot 5^5} (color{red}{2 cdot 5})}$$
$$log_{2^3 cdot 5^5} (x) = frac{1}{log_{2^3 cdot 5^5} ({2 cdot 5})} cdot log_{2 cdot 5} (x)$$
but $log_{2^3 cdot 5^5} ({2 cdot 5}) ne 15$.
answered Nov 18 at 7:07
Toby Mak
3,32811128
edited Nov 18 at 7:12
answered Nov 18 at 7:07
Toby Mak
3,32811128
answered Nov 18 at 7:07
Toby Mak
3,32811128
answered Nov 18 at 7:07
Toby Mak
3,32811128
3,32811128
add a comment |
add a comment |
I cannot read the equation. Is it $2^3*5^5(x)=frac{1}{3*5}text{log}_{(2*5)}(x)$?
– Joey Kilpatrick
Nov 18 at 6:52
@Rohan This is what MathJax looks like. If this is not what you wanted, you can change it yourself.
– Toby Mak
Nov 18 at 6:54
it's correct now
– Rohan
Nov 18 at 7:02