Solve for X: Log 289 Log 17 = Logx

Solve for x: `("log"289)/("log"17)` = logx

Sum

`("log"289)/("log"17)` = logx

⇒ `("log"17^2)/("log"17)` = logx

⇒ `(2"log"17)/("log"17)` = logx
⇒ 2 = logx
⇒ 2log10 = logx ...(since log10 = 1)
⇒ log10² = logx
∴ x = 10²
= 100.

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More About Logarithm

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Chapter 10: Logarithms - Exercise 10.2

Chapter 10 Logarithms
Exercise 10.2 | Q 8.8

If x = 1 + log 2 - log 5, y = 2 log3 and z = log a - log 5; find the value of a if x + y = 2z.

If m = log 20 and n = log 25, find the value of x, so that :
2 log (x - 4) = 2 m - n.

Find x, if : log_x 625 = - 4

Given log₁₀x = 2a and log₁₀y = `b/2. "If" log_10^p = 3a - 2b`, express P in terms of x and y.

If a² = log x , b³ = log y and `a^2/2 - b^3/3` = log c , find c in terms of x and y.

Find x and y, if `("log"x)/("log"5) = ("log"36)/("log"6) = ("log"64)/("log"y)`

Express the following in a form free from logarithm:
3 log x - 2 log y = 2

Express the following in a form free from logarithm:
`2"log" x + 1/2"log" y` = 1

If log (a + 1) = log (4a - 3) - log 3; find a.

If a = log 20 b = log 25 and 2 log (p - 4) = 2a - b, find the value of 'p'.