Difference between revisions of "1980 AHSME Problems/Problem 18"

m (Solution)
 
Line 14: Line 14:
 
<cmath>b^{2c}+b^{2a}=1</cmath>
 
<cmath>b^{2c}+b^{2a}=1</cmath>
 
<cmath>b^{2c}=1-b^{2a}</cmath>
 
<cmath>b^{2c}=1-b^{2a}</cmath>
<cmath>\log_b 1-b^{2a} = 2c</cmath>
+
<cmath>\log_b (1-b^{2a}) = 2c</cmath>
 
<cmath>c=\boxed{\text{(D)} \ \frac 12 \log_b(1-b^{2a})}</cmath>
 
<cmath>c=\boxed{\text{(D)} \ \frac 12 \log_b(1-b^{2a})}</cmath>
  

Latest revision as of 20:45, 18 June 2021

Problem

If $b>1$, $\sin x>0$, $\cos x>0$, and $\log_b \sin x = a$, then $\log_b \cos x$ equals

$\text{(A)} \ 2\log_b(1-b^{a/2}) ~~\text{(B)} \ \sqrt{1-a^2} ~~\text{(C)} \ b^{a^2} ~~\text{(D)} \ \frac 12 \log_b(1-b^{2a}) ~~\text{(E)} \ \text{none of these}$

Solution

\[\log_b \sin x = a\] \[b^a=\sin x\] \[\log_b \cos x=c\] \[b^c=\cos x\] Since $\sin^2x+\cos^2x=1$, \[(b^c)^2+(b^a)^2=1\] \[b^{2c}+b^{2a}=1\] \[b^{2c}=1-b^{2a}\] \[\log_b (1-b^{2a}) = 2c\] \[c=\boxed{\text{(D)} \ \frac 12 \log_b(1-b^{2a})}\]

-aopspandy

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AHSME Problems and Solutions

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