Question

: If log3 (x + 6) = 2, find 'x'.​

Answer 1

hope it's helpful for you!

Answer 2

Solution :

x = 3

Theory :

The Logarithm function is defined as

$\sf\:f(x) =\log_{b}(x)$

where b > 0 and b ≠ 1 and also x >0, reads as log base b of x.

⇒If $\sf\:\log_{b}(a) = x$ ,in exponent form :

$\implies \: b {}^{x} = a$

Step by step explanation :

We have to Find the value of x

$\rm\log_{3}(x+6)=2$

We know that

$\sf\:\log_{b}(a) = x$ ,in exponent form :

$\implies \: b {}^{x} = a$

Thus ,

$\rm\log_{3}(x+6)=2$

$\sf\implies\:(3)^2=x+6$

$\sf\implies\:9=x+6$

$\sf\implies\:x=9-6$

$\sf\implies\:x=3$

Therefore, the value of x is 3

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More About the topic

Rules of Logarithm :

• Basic rules

$\sf\:1)\log(a) + \log(b) = \log(ab)$

$\sf\:2)\log( \frac{a}{b} ) = \log(a) - \log(b)$

$\sf\:3)\log(a) {}^{n} = nlog(a)$

$\sf\:4)\log_{a}(a) = 1$

$\sf\:5)\log_{a}(1)=0$

$\sf\:6)\log_{a{}^{n}}x=\dfrac{1}{n}\times\log_{a}x$

$\sf\:7)\log_{a{}^{n}}x{}^{m}=\dfrac{m}{n}\times\log_{a}$

•Base changing rule

$\sf\:\log_{a}x = \dfrac{\log_{b}x}{\log_{b}a}$