1984 AIME Problems/Problem 5

Problem

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

Solution

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 $4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed {512}$

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

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