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Difference between revisions of "1984 AIME Problems/Problem 5"

 
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== Problem ==
 
== Problem ==
 +
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 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 ==
−
* [[1984 AIME Problems]]
+
*[[Logarithm]]
 +
 
 +
{{AIME box|year=1984|num-b=4|num-a=6}}
 +
* [[AIME Problems and Solutions]]
 +
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
 +
 
 +
[[Category:Intermediate Algebra Problems]]

Latest revision as of 15:10, 16 May 2024

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: \[x + 3y = 5,\] \[y + 3x = 7.\] Now add the two equations together then simplify, we'll get $x+y=3$. So $\log_8{ab} = \log_8{a} + \log_8{b} = 3$, $ab = 8^3 = \boxed{512}$

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

Adding both of the equations, we get \[\log_8{ab} +2\log_4{ab}=12\] Furthermore, we see that $\log_4 {ab}$ is $\frac{3}{2}$ times $\log_8 {ab}.$ Substituting $\log_8 {ab}$ as $x,$ we get $x+3x=12,$ so $x=3.$ Therefore, we have $\log_8 {ab} = 3,$ so $ab= 8^3=\boxed{512}$ ~ math_comb01

Solution 9

Change all equations to base 64. We then get: \[\log_{64}(a^2) + \log_{64}(b^6) = 5\] and \[\log_{64}(b^2) + \log_{64}(a^6) = 7.\]

Using the property \(\log(a) + \log(b) = \log(ab)\), we get: \[\log_{64}(a^2b^6) = 5\] and \[\log_{64}(a^6b^2) = 7.\]

Then: \[a^2b^6 = 64^5\] and \[a^6b^2 = 64^7.\]

Simplifying, we have: \[ab^3 = 8^5\] and \[a^3b = 8^7.\]

Substituting and solving, we get: \[a = 8^2\] and \[b = 8.\]

Then: \[ab = 8^3 = 512.\]

See also

1984 AIME (Problems • Answer Key • Resources)
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