Question

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Answer 1

${\large\bf{\mid{\overline{\underline{To\:Prove:-}}}\mid}}$

• log(base3)x + 2log(basex)3 = 3

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• log change of base formula that is log(base b)a = (log(base x)a/log(base x)b)
• product of rule of log , that is = log a + log b = log(a × b)
• Expenonant Rule of law .. x.log(a) = log(a)^x

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$\red{\textbf{Excellent Question...}} \\ </p><p>\red{\textbf{My Approach in Easiest way..}}$

Step (1) :-----

→ 2.log(x)3 = log(x)3² = log(x)9

Step (2) :-------

[using Change of base formula in Both LHS now we get]

→ log(3)x = [ log(a)x/log(a)3 ]

→ log(x)9 = [ log(a)9/log(a)x ]

Step (3) :-----------

using both as product law we get,

→ log(3)x + log(x)9

→ [ log(a)x/log(a)3 ] × [ log(a)9/log(a)x ]

→ log(a)9/log(a)3

Step (4) :-----------

Again using Exponent Formula in Numerator we get,

→ 3log(a)3/log(a)3

$\bold{\boxed{\large{\boxed{\orange{\small{\boxed{\large{\red{\bold{\: = 3 = RHS}}}}}}}}}} \:$

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$\red\bigstar \: \textbf{Hence Proved}\red\bigstar \:$

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(Hope it Helps you)

#BAL

Answer 2

Answer:

refer to attachment ✌️✌️