$\lg x+54 \log _{x} 10=15$

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\lg x+54 \log _{x} 10=15

Understand Given Equation: First, let's understand the given equation $\lg x + 54 \cdot \log_{x}10 = 15$ . Here, $\lg x$ is the logarithm of $x$ to the base $10$ , which is the same as $\log_{10}x$ . Also, $\log_{x}10$ is the logarithm of $10$ to the base $x$ . We need to solve for $x$ .
Change of Base Formula: We can use the change of base formula for logarithms, which states that $\log_{a}b = \frac{\log_{c}b}{\log_{c}a}$ , to convert $\log_{x}10$ into a logarithm with base $10$ . This gives us $\frac{\log_{10}10}{\log_{10}x}.$
Simplify Logarithmic Expressions: Since $\log_{10}10$ is equal to $1$ (because any log of a number to the same base is $1$ ), we can simplify the expression to $1 / \log_{10}x$ , which is also equal to $1 / \lg x$ .
Rewrite and Simplify Equation: Now we can rewrite the original equation using this simplification: $\lg x + 54 \times \left(\frac{1}{\lg x}\right) = 15$ .
Convert to Quadratic Equation: To solve this equation, we can multiply through by $\lg x$ to get rid of the fraction: $(\lg x)^2 + 54 = 15 \times \lg x$ .
Apply Quadratic Formula: Rearrange the equation to set it to zero: $(\lg x)^2 - 15 \cdot \lg x + 54 = 0$ .
Calculate Discriminant: This is a quadratic equation in terms of $\lg x$ . We can solve for $\lg x$ using the quadratic formula, where $a = 1$ , $b = -15$ , and $c = 54$ .
Solve for $\lg x$ : The quadratic formula is given by $\lg x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Plugging in the values, we get $\lg x = \frac{15 \pm \sqrt{15^2 - 4\cdot 1\cdot 54}}{2\cdot 1}$ .
Simplify Solutions: Calculate the discriminant: $15^2 - 4\cdot1\cdot54 = 225 - 216 = 9$ .
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = (15 \pm \sqrt{9}) / 2$ . This gives us two solutions: $\lg x = (15 + 3) / 2$ or $\lg x = (15 - 3) / 2$ .
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = (15 \pm \sqrt{9}) / 2$ . This gives us two solutions: $\lg x = (15 + 3) / 2$ or $\lg x = (15 - 3) / 2$ .Simplify both solutions: $\lg x = 18 / 2$ or $\lg x = 12 / 2$ , which gives us $\lg x = 9$ or $\lg x = 6$ .
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = (15 \pm \sqrt{9}) / 2$ . This gives us two solutions: $\lg x = (15 + 3) / 2$ or $\lg x = (15 - 3) / 2$ .Simplify both solutions: $\lg x = 18 / 2$ or $\lg x = 12 / 2$ , which gives us $\lg x = 9$ or $\lg x = 6$ .Convert the logarithmic form to exponential form to solve for x: $10^{\lg x} = x$ . This gives us $x = 10^9$ or $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ .
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = \frac{15 \pm \sqrt{9}}{2}$ . This gives us two solutions: $\lg x = \frac{15 + 3}{2}$ or $\lg x = \frac{15 - 3}{2}$ .Simplify both solutions: $\lg x = \frac{18}{2}$ or $\lg x = \frac{12}{2}$ , which gives us $\lg x = 9$ or $\lg x = 6$ .Convert the logarithmic form to exponential form to solve for x: $10^{\lg x} = x$ . This gives us $x = 10^9$ or $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $0$ .We have found two potential solutions for x, but we need to check if both are valid in the original equation. Let's check $x = 10^9$ and $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $0$ in the equation $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $3$ .
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = \frac{15 \pm \sqrt{9}}{2}$ . This gives us two solutions: $\lg x = \frac{15 + 3}{2}$ or $\lg x = \frac{15 - 3}{2}$ .Simplify both solutions: $\lg x = \frac{18}{2}$ or $\lg x = \frac{12}{2}$ , which gives us $\lg x = 9$ or $\lg x = 6$ .Convert the logarithmic form to exponential form to solve for $x$ : $10^{\lg x} = x$ . This gives us $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $0$ or $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $1$ .We have found two potential solutions for $x$ , but we need to check if both are valid in the original equation. Let's check $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $0$ and $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $1$ in the equation $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $5$ .First, check $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $0$ : $\lg x = \frac{15 \pm \sqrt{9}}{2}$ $7$ . This solution is valid.
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = (15 \pm \sqrt{9}) / 2$ . This gives us two solutions: $\lg x = (15 + 3) / 2$ or $\lg x = (15 - 3) / 2$ . Simplify both solutions: $\lg x = 18 / 2$ or $\lg x = 12 / 2$ , which gives us $\lg x = 9$ or $\lg x = 6$ . Convert the logarithmic form to exponential form to solve for $x$ : $10^{\lg x} = x$ . This gives us $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ or $\lg x = (15 \pm \sqrt{9}) / 2$ $1$ . We have found two potential solutions for $x$ , but we need to check if both are valid in the original equation. Let's check $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ and $\lg x = (15 \pm \sqrt{9}) / 2$ $1$ in the equation $\lg x = (15 \pm \sqrt{9}) / 2$ $5$ . First, check $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ : $\lg x = (15 \pm \sqrt{9}) / 2$ $7$ . This solution is valid. Now, check $\lg x = (15 \pm \sqrt{9}) / 2$ $1$ : $\lg x = (15 \pm \sqrt{9}) / 2$ $9$ . This solution is also valid.
Check Validity of Solutions: Now, calculate the two possible solutions for $\lg x$ : $\lg x = (15 \pm \sqrt{9}) / 2$ . This gives us two solutions: $\lg x = (15 + 3) / 2$ or $\lg x = (15 - 3) / 2$ .Simplify both solutions: $\lg x = 18 / 2$ or $\lg x = 12 / 2$ , which gives us $\lg x = 9$ or $\lg x = 6$ .Convert the logarithmic form to exponential form to solve for x: $10^{\lg x} = x$ . This gives us $x = 10^9$ or $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ .We have found two potential solutions for x, but we need to check if both are valid in the original equation. Let's check $x = 10^9$ and $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ in the equation $\lg x = (15 \pm \sqrt{9}) / 2$ $3$ .First, check $x = 10^9$ : $\lg x = (15 \pm \sqrt{9}) / 2$ $5$ . This solution is valid.Now, check $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ : $\lg x = (15 \pm \sqrt{9}) / 2$ $7$ . This solution is also valid.Both $x = 10^9$ and $\lg x = (15 \pm \sqrt{9}) / 2$ $0$ satisfy the original equation, so there are two valid solutions for x.

lg⁡x+54log⁡x10=15 \lg x+54 \log _{x} 10=15 lgx+54logx​10=15

$\lg x+54 \log _{x} 10=15$