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Question

t $a greater than 1$ be a real number. Then the number of roots equation $a to the power of 2 log subscript 2 end subscript invisible function application x end exponent equals 5 plus 4 x to the power of log subscript 2 end subscript invisible function application a end exponent$ has

2
Infinite
0
1

The correct answer is: 1

Given equation can be written as $open parentheses a to the power of log subscript 2 end subscript invisible function application x end exponent close parentheses to the power of 2 end exponent equals 5 plus 4 blank a to the power of log subscript 2 end subscript invisible function application x end exponent$
Let $a to the power of log subscript 2 end subscript invisible function application x end exponent equals t comma$ then the given equation is $t to the power of 2 end exponent minus 4 t minus 5 equals 0. blank$ We get $open parentheses t minus 5 close parentheses open parentheses t plus 1 close parentheses equals 0$
$rightwards double arrow t equals 5 blank o r blank t equals negative 1$ (rejecte(d)
$therefore a to the power of log subscript 2 end subscript invisible function application x end exponent equals 5 rightwards double arrow x to the power of log subscript 2 end subscript invisible function application a end exponent equals 5 rightwards double arrow x equals 5 to the power of log subscript a end subscript invisible function application 2 end exponent$

Related Questions to study

of ordered pairs which satisfy the equation $x to the power of 2 end exponent plus 2 x sin invisible function application open parentheses x y close parentheses plus 1 equals 0$ are $blank$ (where $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

of ordered pairs which satisfy the equation $x to the power of 2 end exponent plus 2 x sin invisible function application open parentheses x y close parentheses plus 1 equals 0$ are $blank$ (where $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

parallel

$i s blank e q u a l blank t o$

$i s blank e q u a l blank t o$

$sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus 1 over denominator 2 x end fraction comma blank$ then $x blank$ must be

$sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus 1 over denominator 2 x end fraction comma blank$ then $x blank$ must be

e equation $cos to the power of 8 end exponent invisible function application x plus b cos to the power of 4 end exponent invisible function application x plus 1 equals 0$ will have a solution if $b$ belongs to

e equation $cos to the power of 8 end exponent invisible function application x plus b cos to the power of 4 end exponent invisible function application x plus 1 equals 0$ will have a solution if $b$ belongs to

parallel

e minimum vertical distance between the graphs of $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

e minimum vertical distance between the graphs of $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

e number of solutions of $2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets$ is

e number of solutions of $2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets$ is

e general solution of $tan invisible function application theta plus tan invisible function application 2 theta plus tan invisible function application 3 theta equals 0$ is

e general solution of $tan invisible function application theta plus tan invisible function application 2 theta plus tan invisible function application 3 theta equals 0$ is

parallel

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

$alpha plus beta plus gamma equals 2 pi comma$ then

$alpha plus beta plus gamma equals 2 pi comma$ then

$i s blank e q u a l blank t o$

$i s blank e q u a l blank t o$

parallel

t area of triangle $A B C$ is $blank open parentheses square root of 3 minus 1 close parentheses divided by 2 comma blank b equals 2$ and $blank c equals open parentheses square root of 3 minus 1 close parentheses$ and $blank angle A$ is acute. The measure of the angle $C$ is

t area of triangle $A B C$ is $blank open parentheses square root of 3 minus 1 close parentheses divided by 2 comma blank b equals 2$ and $blank c equals open parentheses square root of 3 minus 1 close parentheses$ and $blank angle A$ is acute. The measure of the angle $C$ is

$tan invisible function application beta equals 2 sin invisible function application alpha sin invisible function application gamma blank c o s e c blank open parentheses alpha plus gamma close parentheses comma blank t h e n cot invisible function application alpha comma cot invisible function application beta comma cot invisible function application gamma blank$ are in

$tan invisible function application beta equals 2 sin invisible function application alpha sin invisible function application gamma blank c o s e c blank open parentheses alpha plus gamma close parentheses comma blank t h e n cot invisible function application alpha comma cot invisible function application beta comma cot invisible function application gamma blank$ are in

e equation $sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent over denominator 2 x y end fraction$ is possible if

e equation $sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent over denominator 2 x y end fraction$ is possible if

parallel

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