1962 AHSME Problems/Problem 28: Difference between revisions
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==Solution== | ==Solution== | ||
{{ | Taking the base-<math>x</math> logarithm of both sides gives <math>\log_{10}x=\log_x\frac{x^3}{100}</math>. | ||
This simplifies to | |||
<cmath>\log_{10}x=\log_x{x^3} - \log_x{100}</cmath> | |||
<cmath>\log_{10}x+\log_x{100}=3</cmath> | |||
<cmath>\log_{10}x+2 \log_x{10}=3</cmath> | |||
At this point, we substitute <math>u=\log_{10}x</math>. | |||
<cmath>u+\frac2u=3</cmath> | |||
<cmath>u^2+2=3u</cmath> | |||
<cmath>u^2-3u+2=0</cmath> | |||
<cmath>(u-2)(u-1)=0</cmath> | |||
<cmath>u\in\{1, 2\}</cmath> | |||
<cmath>x\in\{10, 100\}</cmath> | |||
The answer is <math>\boxed{\textbf{(D)}}</math>. | |||
Latest revision as of 20:42, 16 April 2014
Problem
The set of 
-values satisfying the equation 
consists of:
Solution
Taking the base- logarithm of both sides gives 
.
This simplifies to
![\[\log_{10}x=\log_x{x^3} - \log_x{100}\]](http://latex.artofproblemsolving.com/6/2/4/624b8ecc2e2e555fa6f29a2b7695141f07005a1d.png)
![\[\log_{10}x+\log_x{100}=3\]](http://latex.artofproblemsolving.com/a/6/6/a665532485f3aec21e921aba6a56b07218c2906f.png)
![\[\log_{10}x+2 \log_x{10}=3\]](http://latex.artofproblemsolving.com/0/1/8/01895caafa5d9fb0338873143db942ffeb218720.png)
At this point, we substitute 
.
![\[u+\frac2u=3\]](http://latex.artofproblemsolving.com/b/2/d/b2d76b88e90f398b6f0224a8ccdb03553fd2d044.png)
![\[u^2+2=3u\]](http://latex.artofproblemsolving.com/3/e/0/3e047a89b061d4032a5c4eedf40ff30529d846fb.png)
![\[u^2-3u+2=0\]](http://latex.artofproblemsolving.com/a/5/8/a582fb5c25824937c2b704758141c896a3df4b87.png)
![\[(u-2)(u-1)=0\]](http://latex.artofproblemsolving.com/2/3/9/2399937c4b5f2ac396b9f2675d6e0b45af49f25d.png)
![\[u\in\{1, 2\}\]](http://latex.artofproblemsolving.com/1/b/4/1b445d499dee30f93ad43550f37a378aebba2247.png)
![\[x\in\{10, 100\}\]](http://latex.artofproblemsolving.com/b/4/4/b4403d1dcfd00a533b1840f229f61f3836de4c1f.png)
The answer is 
.
