# 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 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) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions