Difference between revisions of "1984 AIME Problems/Problem 5"
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Determine the value of <math>ab</math> if <math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math>. | Determine the value of <math>ab</math> if <math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math>. | ||
− | == Solution == | + | == Solution 1 == |
− | Use the [[change of base formula]] to see that <math>\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5</math>; combine [[denominator]]s to find that <math>\frac{\log | + | Use the [[change of base formula]] to see that <math>\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5</math>; combine [[denominator]]s to find that <math>\frac{\log ab^3}{3\log 2} = 5</math>. Doing the same thing with the second equation yields that <math>\frac{\log a^3b}{3\log 2} = 7</math>. This means that <math>\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}</math> and that <math>\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}</math>. If we multiply the two equations together, we get that <math>a^4b^4 = 2^{36}</math>, so taking the fourth root of that, <math>ab = 2^9 = \boxed{512}</math>. |
+ | |||
+ | == Solution 2 == | ||
+ | We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become <math>\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5</math> and <math>\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7</math>. Adding the equations and factoring, we get <math>(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12</math>. Rearranging we see that <math>\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}</math>. Again, we pull exponents out of our logarithms to get <math>\ln ab = \frac{12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2</math>. This means that <math>\frac{\ln ab}{\ln 2} = 9</math>. The left-hand side can be interpreted as a base-2 logarithm, giving us <math>ab = 2^9 = \boxed{512}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | This solution is very similar to the above two, but it utilizes the well-known fact that <math> \log_{m^k}{n^k}= \log_m{n}. </math> Thus, <math> \log_8a+\log_4b^2=5 \Rightarrow | ||
+ | \log_{2^3}{(\sqrt[3]{a})^3} + \log_{2^2}{b^2} = 5 \Rightarrow \log_2{\sqrt[3]{a}} + \log_2{b} = 5 \Rightarrow \log_2{\sqrt[3]{a}b} = 5. </math> Similarly, <math> \log_8b+\log_4a^2=7 \Rightarrow \log_2{\sqrt[3]{b}a} = 7. </math> Adding these two equations, we have <math> \log_2{a^{\frac{4}{3}}b^{\frac{4}{3}}} = 12 \Rightarrow ab = 2^{12\times\frac{3}{4}} = 2^9 = \boxed{512} </math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | We can change everything to a common base, like so: <math>\log_8{a} + \log_8{b^3} = 5,</math> <math>\log_8{b} + \log_8{a^3} = 7.</math> We set the value of <math>\log_8{a}</math> to <math>x</math>, and the value of <math>\log_8{b}</math> to <math>y.</math> Now we have a system of linear equations: <cmath>x + 3y = 5,</cmath> <cmath>y + 3x = 7.</cmath> Now add the two equations together then simplify, we'll get <math>x+y=3</math>. So <math>\log_8{ab} = \log_8{a} + \log_8{b} = 3</math>, <math>ab = 8^3 = \boxed{512}</math> | ||
+ | |||
+ | == Solution 5 == | ||
+ | Add the two equations to get <math>\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12</math>. This can be simplified with the log property <math>\log_n {x}+\log_n {y}=log_n {xy}</math>. Using this, we get <math>\log_8 {ab}+ \log_4 {a^2b^2}=12</math>. Now let <math>\log_8 {ab}=c</math> and <math>\log_4 {a^2b^2}=k</math>. Converting to exponents, we get <math>8^c=ab</math> and <math>4^k=(ab)^2</math>. Sub in the <math>8^c</math> to get <math>k=3c</math>. So now we have that <math>k+c=12</math> and <math>k=3c</math> which gives <math>c=3</math>, <math>k=9</math>. This means <math>\log_4 {a^2b^2}=9</math> so <math>4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed {512}</math> | ||
+ | |||
+ | == Solution 6 == | ||
+ | Add the equations and use the facts that <math>\log{a} +\log{b}=\log{ab} </math> and <math>\log{k^n} =n\log{k} </math> to get | ||
+ | <cmath>\log_8{ab} +2\log_4{ab}=12</cmath> | ||
+ | Now use the change of base identity with base as 2: | ||
+ | <cmath>\dfrac{\log_2{ab}}{\log_2{8}}+\dfrac{2\log_2{ab}}{\log_2{4}}=12</cmath> | ||
+ | Which gives: | ||
+ | <cmath>\frac{4}{3}\log_2{ab}=12</cmath> | ||
+ | Solving gives, <math>\boxed{ab=2^9=512}</math> | ||
+ | |||
+ | == Solution 7 == | ||
+ | By properties of logarithms, we know that <math>\log_8 {a}+ \log_4 {b ^ 2} = \log_2 {a ^ {1/3}}+ \log_2 {b} = 5</math>. | ||
+ | |||
+ | Using the fact that <math>\log_a {b} + \log_a {c} = log_a{b*c}</math>, we get <math>\log_2 {a^{1/3} * b} = 5</math>. | ||
+ | |||
+ | Similarly, we know that <math>\log_2 {a * b^{1/3}} = 7</math>. | ||
+ | |||
+ | From these two equations, we get <math>a^{1/3} * b = 2^5</math> and <math>a * b^{1/3} = 2^7</math>. | ||
+ | |||
+ | Multiply the two equations to get <math>a^{4/3} * b^{4/3} = 2^{12}</math>. Solving, we get that <math>a*b = 2^{12*3/4} = 2^9 = </math><math>\boxed{512}</math>. | ||
+ | |||
+ | == Solution 8 == | ||
+ | Adding both of the equations, we get | ||
+ | <cmath>\log_8{ab} +2\log_4{ab}=12</cmath> | ||
+ | Furthermore, we see that <math>\log_4 {ab}</math> is <math>\frac{3}{2}</math> times <math>\log_8 {ab}.</math> Substituting <math>\log_8 {ab}</math> as <math>x,</math> we get <math>x+3x=12,</math> so <math>x=3.</math> Therefore, we have <math>\log_8 {ab} = 3,</math> so <math>ab= 8^3=\boxed{512}</math> ~ math_comb01 | ||
+ | |||
+ | == Solution 9 == | ||
+ | |||
+ | Change all equations to base 64. We then get: | ||
+ | <cmath> \log_{64}(a^2) + \log_{64}(b^6) = 5 </cmath> | ||
+ | and | ||
+ | <cmath> \log_{64}(b^2) + \log_{64}(a^6) = 7. </cmath> | ||
+ | |||
+ | Using the property \(\log(a) + \log(b) = \log(ab)\), we get: | ||
+ | <cmath> \log_{64}(a^2b^6) = 5 </cmath> | ||
+ | and | ||
+ | <cmath> \log_{64}(a^6b^2) = 7. </cmath> | ||
+ | |||
+ | Then: | ||
+ | <cmath> a^2b^6 = 64^5 </cmath> | ||
+ | and | ||
+ | <cmath> a^6b^2 = 64^7. </cmath> | ||
+ | |||
+ | Simplifying, we have: | ||
+ | <cmath> ab^3 = 8^5 </cmath> | ||
+ | and | ||
+ | <cmath> a^3b = 8^7. </cmath> | ||
+ | |||
+ | Substituting and solving, we get: | ||
+ | <cmath> a = 8^2 </cmath> | ||
+ | and | ||
+ | <cmath> b = 8. </cmath> | ||
+ | |||
+ | Then: | ||
+ | <cmath> ab = 8^3 = 512. </cmath> | ||
== See also == | == See also == | ||
Line 12: | Line 80: | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 16:10, 16 May 2024
Contents
Problem
Determine the value of if
and
.
Solution 1
Use the change of base formula to see that ; combine denominators to find that
. Doing the same thing with the second equation yields that
. This means that
and that
. If we multiply the two equations together, we get that
, so taking the fourth root of that,
.
Solution 2
We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become and
. Adding the equations and factoring, we get
. Rearranging we see that
. Again, we pull exponents out of our logarithms to get
. This means that
. The left-hand side can be interpreted as a base-2 logarithm, giving us
.
Solution 3
This solution is very similar to the above two, but it utilizes the well-known fact that Thus,
Similarly,
Adding these two equations, we have
.
Solution 4
We can change everything to a common base, like so:
We set the value of
to
, and the value of
to
Now we have a system of linear equations:
Now add the two equations together then simplify, we'll get
. So
,
Solution 5
Add the two equations to get . This can be simplified with the log property
. Using this, we get
. Now let
and
. Converting to exponents, we get
and
. Sub in the
to get
. So now we have that
and
which gives
,
. This means
so
Solution 6
Add the equations and use the facts that and
to get
Now use the change of base identity with base as 2:
Which gives:
Solving gives,
Solution 7
By properties of logarithms, we know that .
Using the fact that , we get
.
Similarly, we know that .
From these two equations, we get and
.
Multiply the two equations to get . Solving, we get that
.
Solution 8
Adding both of the equations, we get
Furthermore, we see that
is
times
Substituting
as
we get
so
Therefore, we have
so
~ math_comb01
Solution 9
Change all equations to base 64. We then get:
and
Using the property \(\log(a) + \log(b) = \log(ab)\), we get:
and
Then:
and
Simplifying, we have:
and
Substituting and solving, we get:
and
Then:
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |