Difference between revisions of "1966 AHSME Problems/Problem 12"

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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
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We know that
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<math>2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}</math>.
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We also know that
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<math>8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}</math>.
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There are infinite solutions to the equation <math>2^{12x+15}=2^{12x+15}</math>, so the answer is <math>\boxed{\text{(E)  greater than 3}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:48, 24 March 2023

Problem

The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:

$\text{(A)  zero} \qquad \text{(B)  one} \qquad \text{(C)  two} \qquad \text{(D)  three} \qquad \text{(E)  greater than 3}$

Solution

We know that $2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}$. We also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$. There are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$, so the answer is $\boxed{\text{(E)  greater than 3}}$.

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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