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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>.
  
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==Solutions==
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== Solution 1 ==
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=== 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 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>.
 
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>.
  
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=== Solution 2 ===
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== 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>.
 
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>.
  
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=== Solution 3 ===
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== 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  
 
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>.
 
\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>.
  
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=== Solution 4 ===
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== 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> We multiply the second equation by three and subtract the first equation from the second to get <math>8x = 16,</math> which gives <math>x=2.</math> We substitute our value into the first equation and find that <math>y = 1.</math> Now we know that <math>\log_8{a} = 2,</math> and <math>\log_8{b} = 1,</math> so we find that <math>a = 64,</math> and <math>b=8.</math> Multiplying together gives us <math>ab = \boxed{512}.</math>
 
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> We multiply the second equation by three and subtract the first equation from the second to get <math>8x = 16,</math> which gives <math>x=2.</math> We substitute our value into the first equation and find that <math>y = 1.</math> Now we know that <math>\log_8{a} = 2,</math> and <math>\log_8{b} = 1,</math> so we find that <math>a = 64,</math> and <math>b=8.</math> Multiplying together gives us <math>ab = \boxed{512}.</math>
  
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=== Solution 5 ===
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== 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>
 
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>
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=== Solution 6 ===
+
 
 +
== 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
 
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>
 
<cmath>\log_8{ab} +2\log_4{ab}=12</cmath>
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<cmath>\frac{4}{3}\log_2{ab}=12</cmath>
 
<cmath>\frac{4}{3}\log_2{ab}=12</cmath>
 
Solving gives, <math>\boxed{ab=2^9=512}</math>
 
Solving gives, <math>\boxed{ab=2^9=512}</math>
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=== Solution 7 ===
+
 
 +
== 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>.  
 
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>.  
  
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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>.
 
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>.
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=== Solution 8 ===
+
 
 +
== Solution 8 ==
 
<math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math> adding both we get  
 
<math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math> adding both we get  
 
<cmath>\log_8{ab} +2\log_4{ab}=12</cmath> then we see <math>\log_4 {ab}</math> is 3/2 times <math>\log_8 {ab}</math> putting <math>\log_8 {ab}</math> as x we get x+3x=12 so x=3 and <math>\log_8 {ab}</math> = 3 so ab= <math>8^3</math>=<cmath>\boxed{512}</cmath> ~ math_comb01
 
<cmath>\log_8{ab} +2\log_4{ab}=12</cmath> then we see <math>\log_4 {ab}</math> is 3/2 times <math>\log_8 {ab}</math> putting <math>\log_8 {ab}</math> as x we get x+3x=12 so x=3 and <math>\log_8 {ab}</math> = 3 so ab= <math>8^3</math>=<cmath>\boxed{512}</cmath> ~ math_comb01

Revision as of 04:14, 21 July 2022

Problem

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$.

Solution 1

Use the change of base formula to see that $\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5$; combine denominators to find that $\frac{\log ab^3}{3\log 2} = 5$. Doing the same thing with the second equation yields that $\frac{\log a^3b}{3\log 2} = 7$. This means that $\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}$ and that $\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}$. If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$, so taking the fourth root of that, $ab = 2^9 = \boxed{512}$.

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 $\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5$ and $\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7$. Adding the equations and factoring, we get $(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12$. Rearranging we see that $\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}$. Again, we pull exponents out of our logarithms to get $\ln ab = \frac{12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2$. This means that $\frac{\ln ab}{\ln 2} = 9$. The left-hand side can be interpreted as a base-2 logarithm, giving us $ab = 2^9 = \boxed{512}$.

Solution 3

This solution is very similar to the above two, but it utilizes the well-known fact that $\log_{m^k}{n^k}= \log_m{n}.$ Thus, $\log_8a+\log_4b^2=5 \Rightarrow \log_{2^3}{(\sqrt[3]{a})^3} + \log_{2^2}{b^2} = 5 \Rightarrow \log_2{\sqrt[3]{a}} + \log_2{b} = 5 \Rightarrow \log_2{\sqrt[3]{a}b} = 5.$ Similarly, $\log_8b+\log_4a^2=7 \Rightarrow \log_2{\sqrt[3]{b}a} = 7.$ Adding these two equations, we have $\log_2{a^{\frac{4}{3}}b^{\frac{4}{3}}} = 12 \Rightarrow ab = 2^{12\times\frac{3}{4}} = 2^9 = \boxed{512}$.

Solution 4

We can change everything to a common base, like so: $\log_8{a} + \log_8{b^3} = 5,$ $\log_8{b} + \log_8{a^3} = 7.$ We set the value of $\log_8{a}$ to $x$, and the value of $\log_8{b}$ to $y.$ Now we have a system of linear equations: \[x + 3y = 5,\] \[y + 3x = 7.\] We multiply the second equation by three and subtract the first equation from the second to get $8x = 16,$ which gives $x=2.$ We substitute our value into the first equation and find that $y = 1.$ Now we know that $\log_8{a} = 2,$ and $\log_8{b} = 1,$ so we find that $a = 64,$ and $b=8.$ Multiplying together gives us $ab = \boxed{512}.$

Solution 5

Add the two equations to get $\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12$. This can be simplified with the log property $\log_n {x}+\log_n {y}=log_n {xy}$. Using this, we get $\log_8 {ab}+ \log_4 {a^2b^2}=12$. Now let $\log_8 {ab}=c$ and $\log_4 {a^2b^2}=k$. Converting to exponents, we get $8^c=ab$ and $4^k=(ab)^2$. Sub in the $8^c$ to get $k=3c$. So now we have that $k+c=12$ and $k=3c$ which gives $c=3$, $k=9$. This means $\log_4 {a^2b^2}=9$ so $4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed {512}$

Solution 6

Add the equations and use the facts that $\log{a} +\log{b}=\log{ab}$ and $\log{k^n} =n\log{k}$ to get \[\log_8{ab} +2\log_4{ab}=12\] Now use the change of base identity with base as 2: \[\dfrac{\log_2{ab}}{\log_2{8}}+\dfrac{2\log_2{ab}}{\log_2{4}}=12\] Which gives: \[\frac{4}{3}\log_2{ab}=12\] Solving gives, $\boxed{ab=2^9=512}$

Solution 7

By properties of logarithms, we know that $\log_8 {a}+ \log_4 {b ^ 2} = \log_2 {a ^ {1/3}}+ \log_2 {b} = 5$.

Using the fact that $\log_a {b} + \log_a {c} = log_a{b*c}$, we get $\log_2 {a^{1/3} * b} = 5$.

Similarly, we know that $\log_2 {a * b^{1/3}} = 7$.

From these two equations, we get $a^{1/3} * b = 2^5$ and $a * b^{1/3} = 2^7$.

Multiply the two equations to get $a^{4/3} * b^{4/3} = 2^{12}$. Solving, we get that $a*b = 2^{12*3/4} = 2^9 =$$\boxed{512}$.

Solution 8

$\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ adding both we get \[\log_8{ab} +2\log_4{ab}=12\] then we see $\log_4 {ab}$ is 3/2 times $\log_8 {ab}$ putting $\log_8 {ab}$ as x we get x+3x=12 so x=3 and $\log_8 {ab}$ = 3 so ab= $8^3$=\[\boxed{512}\] ~ math_comb01

See also

1984 AIME (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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