JEE(ADVANCED)-201

LEVELJEE Main

Animated Solution for Mathematics - Basic Mathematics: The value of $((lo g_{2} 9)^{2})^{\frac{1}{l o g _{2} ( l o g _{2} 9 )}} \times (7)^{\frac{1}{l o g _{4} 7}}$ is ________.

Q: Why can't I just calculate the value of $\log_2 9$ directly?

Great question! $\log_2 9$ is an irrational number. If you try to calculate its decimal value, you will get lost in approximations. In JEE problems, always treat such terms as variables (like $A$) and use logarithmic properties to simplify the expression algebraically. The values will often cancel out perfectly.

Enter Numerical Value:

Visualized Solution

The Given Expression $E$

Let the given expression be $E = T_{1} \times T_{2}$
$T_{1} = ((lo g_{2} 9)^{2})^{\frac{1}{l o g _{2} ( l o g _{2} 9 )}}$
$T_{2} = (7)^{\frac{1}{l o g _{4} 7}}$

Substitution in $T_{1}$

Let $A = lo g_{2} 9$
Substitute $A$ into $T_{1}$ :
$T_{1} = (A^{2})^{\frac{1}{l o g _{2} A}}$

Applying $(x^{y})^{z} = x^{y z}$

We know that $(x^{y})^{z} = x^{y z}$
Applying this to $T_{1}$ :
$T_{1} = A^{2 \times \frac{1}{l o g _{2} A}} = A^{\frac{2}{l o g _{2} A}}$

Base Change Formula

Recall the base change property: $\frac{1}{l o g _{a} b} = lo g_{b} a$
We have the term $\frac{1}{l o g _{2} A}$ in the exponent.
Applying the property: $\frac{1}{l o g _{2} A} = lo g_{A} 2$

Applying $c lo g_{a} b = lo g_{a} (b^{c})$

Substitute back into $T_{1}$ :
$T_{1} = A^{2 l o g_{A} 2}$
Recall the power rule for logarithms: $c lo g_{a} b = lo g_{a} (b^{c})$
Apply this to the exponent $2 lo g_{A} 2$ :
$2 lo g_{A} 2 = lo g_{A} (2^{2}) = lo g_{A} 4$

Evaluating $T_{1}$

Now, $T_{1} = A^{l o g_{A} 4}$
Recall the fundamental identity: $a^{l o g_{a} x} = x$
Applying the identity to $T_{1}$ :
The base $A$ and the log base $A$ cancel out.
Therefore, $T_{1} = 4$

Analyzing $T_{2}$

Now let's evaluate $T_{2} = (7)^{\frac{1}{l o g _{4} 7}}$
Rewrite the square root as a fractional power:
$7 = 7^{\frac{1}{2}}$
$T_{2} = (7^{\frac{1}{2}})^{\frac{1}{l o g _{4} 7}}$

Base Change in $T_{2}$

Again, use the base change property: $\frac{1}{l o g _{a} b} = lo g_{b} a$
Apply to $\frac{1}{l o g _{4} 7}$ :
$\frac{1}{l o g _{4} 7} = lo g_{7} 4$

Multiplying Powers in $T_{2}$

Substitute the simplified exponent back:
$T_{2} = (7^{\frac{1}{2}})^{l o g_{7} 4}$
Multiply the powers:
$T_{2} = 7^{\frac{1}{2} l o g_{7} 4}$

Log Power Rule in $T_{2}$

Apply the power rule $c lo g_{a} b = lo g_{a} (b^{c})$ to the exponent:
$\frac{1}{2} lo g_{7} 4 = lo g_{7} (4^{\frac{1}{2}})$
Since $4^{\frac{1}{2}} = 4 = 2$ :
The exponent becomes $lo g_{7} 2$

Evaluating $T_{2}$

Now, $T_{2} = 7^{l o g_{7} 2}$
Using the fundamental identity $a^{l o g_{a} x} = x$ :
$T_{2} = 2$

Final Calculation of $E$

We have found $T_{1} = 4$ and $T_{2} = 2$
The original expression is $E = T_{1} \times T_{2}$
$E = 4 \times 2 = 8$
Final Answer: 8

00:00 / 00:00

The Sigma Insight: Properties of Logarithms

Analyzing the Setup

The given expression is:

E = ((lo g_{2} 9)^{2})^{\frac{1}{l o g _{2} ( l o g _{2} 9 )}} \times (7)^{\frac{1}{l o g _{4} 7}}

To simplify this, we split the expression into two manageable parts:

T_{1}

and

T_{2}

Taming the First Beast

Let

T_{1} = ((lo g_{2} 9)^{2})^{\frac{1}{l o g _{2} ( l o g _{2} 9 )}}

. To simplify, we substitute

A = lo g_{2} 9

The expression becomes:

T_{1} = (A^{2})^{\frac{1}{l o g _{2} A}}

Using the power rule

(x^{y})^{z} = x^{y z}

, we rewrite this as:

T_{1} = A^{\frac{2}{l o g _{2} A}}

Applying the base change property

\frac{1}{l o g _{a} b} = lo g_{b} a

, we transform the exponent:

T_{1} = A^{2 l o g_{A} 2}

Using the power rule

c lo g_{a} b = lo g_{a} (b^{c})

, we move the coefficient

2

inside the logarithm:

T_{1} = A^{l o g_{A} (2^{2})} = A^{l o g_{A} 4}

Applying the fundamental identity

a^{l o g_{a} x} = x

, we find: $T_{1} = 4$

The Second Challenge

Now, consider

T_{2} = (7)^{\frac{1}{l o g _{4} 7}}

. We express the base as

7^{1/2}

T_{2} = (7^{1/2})^{\frac{1}{l o g _{4} 7}}

Applying the base change property

\frac{1}{l o g _{4} 7} = lo g_{7} 4

, the expression becomes:

T_{2} = (7^{1/2})^{l o g_{7} 4}

Multiplying the exponents, we get:

T_{2} = 7^{\frac{1}{2} l o g_{7} 4}

Using the power rule

\frac{1}{2} lo g_{7} 4 = lo g_{7} (4^{1/2}) = lo g_{7} 2

, we simplify further:

T_{2} = 7^{l o g_{7} 2}

Applying the fundamental identity

a^{l o g_{a} x} = x

, we find: $T_{2} = 2$

Final Calculation

We have successfully reduced the complex expression to the product of two simple integers.

The total value is:

E = T_{1} \times T_{2} = 4 \times 2

$E = 8$

Common Traps & FAQs

Why can't I just calculate the value of

lo g_{2} 9

directly?

Great question!

lo g_{2} 9

is an irrational number. If you try to calculate its decimal value, you will get lost in approximations. In JEE problems, always treat such terms as variables (like

A

) and use logarithmic properties to simplify the expression algebraically. The values will often cancel out perfectly.

How do I know when to use the base change formula?

Look at the exponent. If the base of the logarithm in the exponent does not match the base of the term being raised, that is your signal. The base change formula

\frac{1}{l o g _{a} b} = lo g_{b} a

is designed to flip the logarithm, which often allows the bases to align, enabling the use of the fundamental identity

a^{l o g_{a} x} = x

Is there a shortcut for

a^{l o g_{a} x}

It is not just a shortcut; it is a fundamental definition of a logarithm! The expression

lo g_{a} x

asks the question: 'To what power must we raise

a

to get

x

?' Therefore, raising

a

to that power must, by definition, give you

x

. Recognizing this identity is the key to solving many complex-looking logarithmic problems.

Animated Solution for Mathematics - Basic Mathematics: The value of $((lo g_{2} 9)^{2})^{\frac{1}{l o g _{2} ( l o g _{2} 9 )}} \times (7)^{\frac{1}{l o g _{4} 7}}$ is ________.

Enter Numerical Value:

Visualized Solution

The Given Expression $E$

Substitution in $T_{1}$

Applying $(x^{y})^{z} = x^{y z}$

Base Change Formula

Applying $c lo g_{a} b = lo g_{a} (b^{c})$

Evaluating $T_{1}$

Analyzing $T_{2}$

Base Change in $T_{2}$

Multiplying Powers in $T_{2}$

Log Power Rule in $T_{2}$

Evaluating $T_{2}$

Final Calculation of $E$

The Sigma Insight: Properties of Logarithms

Analyzing the Setup

Taming the First Beast

The Second Challenge

Final Calculation

Similar Questions

Let $a = 32$ and $b = \frac{1}{5 ^{1/6} 6}$ . If $x, y \in R$ are such that $3 x + 2 y = lo g_{a} (18)^{5/4}$ and $2 x - y = lo g_{b} (1080)$ , then $4 x + 5 y$ is equal to ________.

Let $a, b, c$ be the three distinct positive real numbers such that $(2 a)^{l o g_{e} a} = (b c)^{l o g_{e} b}$ and $b^{l o g_{e} 2} = a^{l o g_{e} c}$ . Then $6 a + 5 b c$ is equal to ______.

The least value of the expression $2 lo g_{10} x - lo g_{x} (0.01)$ , for $x > 1$ , is

The number $lo g_{2} 7$ is

If $n$ is a natural number such that $n = p_{1}^{α_{1}} \cdot p_{2}^{α_{2}} \cdot p_{3}^{α_{3}} \dots p_{k}^{α_{k}}$ and $p_{1}, p_{2}, \dots p_{k}$ are distinct primes, then show that $ln n \geq k ln 2$ .

Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ((log2​9)2)log2​(log2​9)1​×(7​)log4​71​ is ________.

Enter Numerical Value:

Visualized Solution

The Given Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Substitution in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (xy)z=xyz

Base Change Formula

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cloga​b=loga​(bc)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Base Change in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Multiplying Powers in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Log Power Rule in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

The Sigma Insight: Properties of Logarithms

Analyzing the Setup

Taming the First Beast

The Second Challenge

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=32​ and b=51/66​1​. If x,y∈R are such that 3x+2y=loga​(18)5/4 and 2x−y=logb​(1080​), then 4x+5y is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be the three distinct positive real numbers such that (2a)loge​a=(bc)loge​b and bloge​2=aloge​c. Then 6a+5bc is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least value of the expression 2log10​x−logx​(0.01), for x>1, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number log2​7 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n is a natural number such that n=p1α1​​⋅p2α2​​⋅p3α3​​…pkαk​​ and p1​,p2​,…pk​ are distinct primes, then show that lnn≥kln2.

Animated Solution for Mathematics - Basic Mathematics: The value of $((lo g_{2} 9)^{2})^{\frac{1}{l o g _{2} ( l o g _{2} 9 )}} \times (7)^{\frac{1}{l o g _{4} 7}}$ is ________.

The Given Expression $E$

Substitution in $T_{1}$

Applying $(x^{y})^{z} = x^{y z}$

Applying $c lo g_{a} b = lo g_{a} (b^{c})$

Evaluating $T_{1}$

Analyzing $T_{2}$

Base Change in $T_{2}$

Multiplying Powers in $T_{2}$

Log Power Rule in $T_{2}$

Evaluating $T_{2}$

Final Calculation of $E$

Let $a = 32$ and $b = \frac{1}{5 ^{1/6} 6}$ . If $x, y \in R$ are such that $3 x + 2 y = lo g_{a} (18)^{5/4}$ and $2 x - y = lo g_{b} (1080)$ , then $4 x + 5 y$ is equal to ________.

Let $a, b, c$ be the three distinct positive real numbers such that $(2 a)^{l o g_{e} a} = (b c)^{l o g_{e} b}$ and $b^{l o g_{e} 2} = a^{l o g_{e} c}$ . Then $6 a + 5 b c$ is equal to ______.

The least value of the expression $2 lo g_{10} x - lo g_{x} (0.01)$ , for $x > 1$ , is

The number $lo g_{2} 7$ is

If $n$ is a natural number such that $n = p_{1}^{α_{1}} \cdot p_{2}^{α_{2}} \cdot p_{3}^{α_{3}} \dots p_{k}^{α_{k}}$ and $p_{1}, p_{2}, \dots p_{k}$ are distinct primes, then show that $ln n \geq k ln 2$ .