Let a = 3√2 and b = 1/(5^(1/6)√6). If x, y ∈ R are such that 3x+ 2y = log_a(18)^(5/4) and 2x - y = l

Q: Why did we convert the base $b$ to negative exponents? Isn't that making it more complicated?

It is actually the opposite! By bringing everything to the numerator, we can easily compare exponents of prime factors like $5$ and $6$ on both sides of the equation $b^p = A$. It turns a logarithmic problem into a simple algebraic comparison of exponents, which is much less prone to error.

Q: How did you know $1080$ was $6^3 \cdot 5$ so quickly?

Prime factorization is a superpower in JEE. Whenever you see a large number in a log argument, break it down into its prime building blocks ($2, 3, 5$). $1080 = 108 \times 10 = (6^2 \times 3) \times (2 \times 5) = 6^2 \times 6 \times 5 = 6^3 \times 5$. Practice makes this intuitive!

Q: Can I solve for $x$ and $y$ separately, or is there a shortcut for $4x + 5y$?

While you could look for a linear combination of the two equations to get $4x + 5y$ directly, solving for $x$ and $y$ individually is safer and less prone to arithmetic errors. In this case, the system is simple enough that solving for $x$ and $y$ takes less than a minute.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the First Equation

We begin with the equation $3x + 2y = \log_{a}(18)^{5/4}$, where the base is $a = 3\sqrt{2}$.

First, observe that squaring the base yields:
$$a^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18$$

Using the power rule $\log(m^n) = n \cdot \log(m)$, we simplify the expression:
$$3x + 2y = \frac{5}{4} \log_{a}(18)$$

Since $18 = a^2$, the expression becomes $\frac{5}{4} \cdot 2 = \frac{5}{2}$. Thus, our first linear equation is:
$$3x + 2y = \frac{5}{2}$$

### The Prime Factorization Adventure

Next, we address the second equation: $2x - y = \log_{b}(\sqrt{1080})$. The base is given as $b = \frac{1}{5^{1/6}\sqrt{6}}$, which we rewrite as $b = 5^{-1/6} \cdot 6^{-1/2}$.

We factorize the argument $\sqrt{1080}$ as follows:
$$1080 = 108 \cdot 10 = (6^2 \cdot 3) \cdot (2 \cdot 5) = 6^3 \cdot 5$$
$$\sqrt{1080} = (6^3 \cdot 5)^{1/2} = 6^{3/2} \cdot 5^{1/2}$$

We seek $p$ such that $b^p = 6^{3/2} \cdot 5^{1/2}$. Substituting the expression for $b$:
$$(5^{-1/6} \cdot 6^{-1/2})^p = 6^{3/2} \cdot 5^{1/2}$$

Comparing the exponents of $6$, we find $-\frac{p}{2} = \frac{3}{2}$, which implies $p = -3$. Therefore, the second equation is:
$$2x - y = -3$$

### The Final Victory

We now solve the system of linear equations:
1) $3x + 2y = \frac{5}{2}$
2) $2x - y = -3$

Multiply the second equation by $2$ to obtain $4x - 2y = -6$. Adding this to the first equation eliminates $y$:
$$7x = \frac{5}{2} - 6 = -\frac{7}{2}$$
$$x = -\frac{1}{2}$$

Substituting $x = -\frac{1}{2}$ into $2x - y = -3$, we find:
$$2(-\frac{1}{2}) - y = -3 \Rightarrow -1 - y = -3 \Rightarrow y = 2$$

Finally, we calculate the requested value:
$$4x + 5y = 4\left(-\frac{1}{2}\right) + 5(2) = -2 + 10 = 8$$

The final answer is **8**.

Animated Solution for Mathematics - Basic Mathematics: 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 ________.

Enter Numerical Value:

Visualized Solution

Analyze the Given Equations

Simplify $lo g_{a} (18)^{5/4}$

Evaluate Base $a$ and Argument $18$

Analyze $lo g_{b} (1080)$

Factorize $1080$

Express Base $b$ with Exponents

Find the Power $p$

Solve the Linear System

Calculate $4 x + 5 y$

The Sigma Insight: Properties of Logarithms

Analyzing the First Equation

The Prime Factorization Adventure

The Final Victory

Similar Questions

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 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 least value of the expression $2 lo g_{10} x - lo g_{x} (0.01)$ , for $x > 1$ , 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 $ln n \geq k ln 2$ .

The number $lo g_{2} 7$ is

Enter Numerical Value:

Visualized Solution

Analyze the Given Equations

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } loga​(18)5/4

Evaluate Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and Argument .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 18

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logb​(1080​)

Factorize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1080

Express Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b with Exponents

Find the Power .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p