How to evaluate $45^frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm?
up vote
1
down vote
favorite
Let $90^a=2$ and $90^b=5$, Evaluate
$45^frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
logarithms exponentiation
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
|
show 1 more comment
up vote
1
down vote
favorite
Let $90^a=2$ and $90^b=5$, Evaluate
$45^frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
logarithms exponentiation
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
1
I think something along the lines of $10^{-1} = 90^{-a-b} implies 3^2 = 90^{1-a-b} implies 3 = 90^{frac{1-a-b}{2}}$. I'll put further information as I move along.
– hjpotter92
Sep 4 '15 at 9:23
Maybe along the lines of letting $sqrt[a]{2}=90; , sqrt[b]{5}=90$ and knowing that $45 times 2 = 90$ plus also use @hjpotter92 's comment.
– Aldon
Sep 4 '15 at 9:29
@hjpotter92 Thank you for your help. I have a question though. Can you show me how $3^2$ is $90^{1-a-b}$?
– rainbowbutterunikitteh
Sep 4 '15 at 9:32
@Aldon thanks for your input :)
– rainbowbutterunikitteh
Sep 4 '15 at 9:33
@DioPauloHonorario Multiplying both sides by $90$ in $10^{-1} = 90^{-a-b}$
– hjpotter92
Sep 4 '15 at 9:38
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $90^a=2$ and $90^b=5$, Evaluate
$45^frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
logarithms exponentiation
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
Let $90^a=2$ and $90^b=5$, Evaluate
$45^frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
logarithms exponentiation
logarithms exponentiation
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
asked Sep 4 '15 at 9:14
rainbowbutterunikitteh
5116
5116
1
I think something along the lines of $10^{-1} = 90^{-a-b} implies 3^2 = 90^{1-a-b} implies 3 = 90^{frac{1-a-b}{2}}$. I'll put further information as I move along.
– hjpotter92
Sep 4 '15 at 9:23
Maybe along the lines of letting $sqrt[a]{2}=90; , sqrt[b]{5}=90$ and knowing that $45 times 2 = 90$ plus also use @hjpotter92 's comment.
– Aldon
Sep 4 '15 at 9:29
@hjpotter92 Thank you for your help. I have a question though. Can you show me how $3^2$ is $90^{1-a-b}$?
– rainbowbutterunikitteh
Sep 4 '15 at 9:32
@Aldon thanks for your input :)
– rainbowbutterunikitteh
Sep 4 '15 at 9:33
@DioPauloHonorario Multiplying both sides by $90$ in $10^{-1} = 90^{-a-b}$
– hjpotter92
Sep 4 '15 at 9:38
|
show 1 more comment
1
I think something along the lines of $10^{-1} = 90^{-a-b} implies 3^2 = 90^{1-a-b} implies 3 = 90^{frac{1-a-b}{2}}$. I'll put further information as I move along.
– hjpotter92
Sep 4 '15 at 9:23
Maybe along the lines of letting $sqrt[a]{2}=90; , sqrt[b]{5}=90$ and knowing that $45 times 2 = 90$ plus also use @hjpotter92 's comment.
– Aldon
Sep 4 '15 at 9:29
@hjpotter92 Thank you for your help. I have a question though. Can you show me how $3^2$ is $90^{1-a-b}$?
– rainbowbutterunikitteh
Sep 4 '15 at 9:32
@Aldon thanks for your input :)
– rainbowbutterunikitteh
Sep 4 '15 at 9:33
@DioPauloHonorario Multiplying both sides by $90$ in $10^{-1} = 90^{-a-b}$
– hjpotter92
Sep 4 '15 at 9:38
1
1
I think something along the lines of $10^{-1} = 90^{-a-b} implies 3^2 = 90^{1-a-b} implies 3 = 90^{frac{1-a-b}{2}}$. I'll put further information as I move along.
– hjpotter92
Sep 4 '15 at 9:23
I think something along the lines of $10^{-1} = 90^{-a-b} implies 3^2 = 90^{1-a-b} implies 3 = 90^{frac{1-a-b}{2}}$. I'll put further information as I move along.
– hjpotter92
Sep 4 '15 at 9:23
Maybe along the lines of letting $sqrt[a]{2}=90; , sqrt[b]{5}=90$ and knowing that $45 times 2 = 90$ plus also use @hjpotter92 's comment.
– Aldon
Sep 4 '15 at 9:29
Maybe along the lines of letting $sqrt[a]{2}=90; , sqrt[b]{5}=90$ and knowing that $45 times 2 = 90$ plus also use @hjpotter92 's comment.
– Aldon
Sep 4 '15 at 9:29
@hjpotter92 Thank you for your help. I have a question though. Can you show me how $3^2$ is $90^{1-a-b}$?
– rainbowbutterunikitteh
Sep 4 '15 at 9:32
@hjpotter92 Thank you for your help. I have a question though. Can you show me how $3^2$ is $90^{1-a-b}$?
– rainbowbutterunikitteh
Sep 4 '15 at 9:32
@Aldon thanks for your input :)
– rainbowbutterunikitteh
Sep 4 '15 at 9:33
@Aldon thanks for your input :)
– rainbowbutterunikitteh
Sep 4 '15 at 9:33
@DioPauloHonorario Multiplying both sides by $90$ in $10^{-1} = 90^{-a-b}$
– hjpotter92
Sep 4 '15 at 9:38
@DioPauloHonorario Multiplying both sides by $90$ in $10^{-1} = 90^{-a-b}$
– hjpotter92
Sep 4 '15 at 9:38
|
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
Let me try.
$$10 = 90^{a+b} Rightarrow 3^2 = 90^{1-a-b} Rightarrow 3 = 90^{frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{frac{1}{1-a}} = 90 Rightarrow 45^{frac{1-a-b}{2(1-a)}} = 90^{frac{1-a-b}{2}} = 3.$$
answered Sep 4 '15 at 9:34
GAVD
6,64611129
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
add a comment |
up vote
0
down vote
$$2=90^a=(2cdot5cdot3^2)^aiff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^biff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2cdot5)^{1-a-b}=3^{2-2(a+b)}cdot5^{1-a-b}=3^{2-2(a+b)}cdot3^{2b}$$
$$implies45^{1-a-b}=3^{2(1-a)}$$
$implies$ one of the values of $$45^{frac{1-a-b}{2(1-a)}}$$ is $3$
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
add a comment |
up vote
-1
down vote
Given $90^a = 2$, $90^b = 5$
$$(90/2)^{(1-a-b)/(2–2a)}$$
$$=(90/90^a)^{(1-a-b)/(2–2a)}$$
$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$
$$=90^{(1-a-b)/2}$$
$$=sqrt{frac{90}{90^a 90^b}}$$
$$=sqrt{frac{90}{2cdot 5}} = sqrt{9} =3$$
edited Dec 4 at 14:24
jgon
11.6k21839
answered Dec 4 at 13:28
JUMAR VELASCO
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Let me try.
$$10 = 90^{a+b} Rightarrow 3^2 = 90^{1-a-b} Rightarrow 3 = 90^{frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{frac{1}{1-a}} = 90 Rightarrow 45^{frac{1-a-b}{2(1-a)}} = 90^{frac{1-a-b}{2}} = 3.$$
answered Sep 4 '15 at 9:34
GAVD
6,64611129
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
add a comment |
up vote
5
down vote
accepted
Let me try.
$$10 = 90^{a+b} Rightarrow 3^2 = 90^{1-a-b} Rightarrow 3 = 90^{frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{frac{1}{1-a}} = 90 Rightarrow 45^{frac{1-a-b}{2(1-a)}} = 90^{frac{1-a-b}{2}} = 3.$$
answered Sep 4 '15 at 9:34
GAVD
6,64611129
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let me try.
$$10 = 90^{a+b} Rightarrow 3^2 = 90^{1-a-b} Rightarrow 3 = 90^{frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{frac{1}{1-a}} = 90 Rightarrow 45^{frac{1-a-b}{2(1-a)}} = 90^{frac{1-a-b}{2}} = 3.$$
answered Sep 4 '15 at 9:34
GAVD
6,64611129
Let me try.
$$10 = 90^{a+b} Rightarrow 3^2 = 90^{1-a-b} Rightarrow 3 = 90^{frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{frac{1}{1-a}} = 90 Rightarrow 45^{frac{1-a-b}{2(1-a)}} = 90^{frac{1-a-b}{2}} = 3.$$
answered Sep 4 '15 at 9:34
GAVD
6,64611129
answered Sep 4 '15 at 9:34
GAVD
6,64611129
answered Sep 4 '15 at 9:34
GAVD
6,64611129
answered Sep 4 '15 at 9:34
GAVD
6,64611129
6,64611129
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
add a comment |
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
I had just arrived at the result $45=90^{1-a}$. =)
– hjpotter92
Sep 4 '15 at 9:38
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
But your comments are very helpful for OP :)
– GAVD
Sep 4 '15 at 9:43
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
@GAVD Thank you so much for making my life simpler! :))
– rainbowbutterunikitteh
Sep 4 '15 at 9:52
add a comment |
up vote
0
down vote
$$2=90^a=(2cdot5cdot3^2)^aiff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^biff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2cdot5)^{1-a-b}=3^{2-2(a+b)}cdot5^{1-a-b}=3^{2-2(a+b)}cdot3^{2b}$$
$$implies45^{1-a-b}=3^{2(1-a)}$$
$implies$ one of the values of $$45^{frac{1-a-b}{2(1-a)}}$$ is $3$
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
add a comment |
up vote
0
down vote
$$2=90^a=(2cdot5cdot3^2)^aiff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^biff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2cdot5)^{1-a-b}=3^{2-2(a+b)}cdot5^{1-a-b}=3^{2-2(a+b)}cdot3^{2b}$$
$$implies45^{1-a-b}=3^{2(1-a)}$$
$implies$ one of the values of $$45^{frac{1-a-b}{2(1-a)}}$$ is $3$
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
add a comment |
up vote
0
down vote
up vote
0
down vote
$$2=90^a=(2cdot5cdot3^2)^aiff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^biff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2cdot5)^{1-a-b}=3^{2-2(a+b)}cdot5^{1-a-b}=3^{2-2(a+b)}cdot3^{2b}$$
$$implies45^{1-a-b}=3^{2(1-a)}$$
$implies$ one of the values of $$45^{frac{1-a-b}{2(1-a)}}$$ is $3$
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
$$2=90^a=(2cdot5cdot3^2)^aiff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^biff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2cdot5)^{1-a-b}=3^{2-2(a+b)}cdot5^{1-a-b}=3^{2-2(a+b)}cdot3^{2b}$$
$$implies45^{1-a-b}=3^{2(1-a)}$$
$implies$ one of the values of $$45^{frac{1-a-b}{2(1-a)}}$$ is $3$
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
answered Sep 4 '15 at 10:58
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
-1
down vote
Given $90^a = 2$, $90^b = 5$
$$(90/2)^{(1-a-b)/(2–2a)}$$
$$=(90/90^a)^{(1-a-b)/(2–2a)}$$
$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$
$$=90^{(1-a-b)/2}$$
$$=sqrt{frac{90}{90^a 90^b}}$$
$$=sqrt{frac{90}{2cdot 5}} = sqrt{9} =3$$
edited Dec 4 at 14:24
jgon
11.6k21839
answered Dec 4 at 13:28
JUMAR VELASCO
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
add a comment |
up vote
-1
down vote
Given $90^a = 2$, $90^b = 5$
$$(90/2)^{(1-a-b)/(2–2a)}$$
$$=(90/90^a)^{(1-a-b)/(2–2a)}$$
$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$
$$=90^{(1-a-b)/2}$$
$$=sqrt{frac{90}{90^a 90^b}}$$
$$=sqrt{frac{90}{2cdot 5}} = sqrt{9} =3$$
edited Dec 4 at 14:24
jgon
11.6k21839
answered Dec 4 at 13:28
JUMAR VELASCO
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Given $90^a = 2$, $90^b = 5$
$$(90/2)^{(1-a-b)/(2–2a)}$$
$$=(90/90^a)^{(1-a-b)/(2–2a)}$$
$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$
$$=90^{(1-a-b)/2}$$
$$=sqrt{frac{90}{90^a 90^b}}$$
$$=sqrt{frac{90}{2cdot 5}} = sqrt{9} =3$$
edited Dec 4 at 14:24
jgon
11.6k21839
answered Dec 4 at 13:28
JUMAR VELASCO
1
Given $90^a = 2$, $90^b = 5$
$$(90/2)^{(1-a-b)/(2–2a)}$$
$$=(90/90^a)^{(1-a-b)/(2–2a)}$$
$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$
$$=90^{(1-a-b)/2}$$
$$=sqrt{frac{90}{90^a 90^b}}$$
$$=sqrt{frac{90}{2cdot 5}} = sqrt{9} =3$$
edited Dec 4 at 14:24
jgon
11.6k21839
answered Dec 4 at 13:28
JUMAR VELASCO
1
edited Dec 4 at 14:24
jgon
11.6k21839
edited Dec 4 at 14:24
jgon
11.6k21839
edited Dec 4 at 14:24
jgon
11.6k21839
11.6k21839
answered Dec 4 at 13:28
JUMAR VELASCO
1
answered Dec 4 at 13:28
JUMAR VELASCO
1
answered Dec 4 at 13:28
JUMAR VELASCO
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 4 at 13:48
add a comment |
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I think something along the lines of $10^{-1} = 90^{-a-b} implies 3^2 = 90^{1-a-b} implies 3 = 90^{frac{1-a-b}{2}}$. I'll put further information as I move along.
– hjpotter92
Sep 4 '15 at 9:23
Maybe along the lines of letting $sqrt[a]{2}=90; , sqrt[b]{5}=90$ and knowing that $45 times 2 = 90$ plus also use @hjpotter92 's comment.
– Aldon
Sep 4 '15 at 9:29
@hjpotter92 Thank you for your help. I have a question though. Can you show me how $3^2$ is $90^{1-a-b}$?
– rainbowbutterunikitteh
Sep 4 '15 at 9:32
@Aldon thanks for your input :)
– rainbowbutterunikitteh
Sep 4 '15 at 9:33
@DioPauloHonorario Multiplying both sides by $90$ in $10^{-1} = 90^{-a-b}$
– hjpotter92
Sep 4 '15 at 9:38