The value of the expression antilog ( log (1/27). antilog . log (1/10) 7) is in the form of √[b]a, where a, b ∈ N then find the least value of (a+b-2).

Question

The value of the expression antilog ( log (1/27). antilog . log (1/10) 7) is in the form of √[b]a, where a, b ∈ N then find the least value of (a+b-2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

( log (1/27) operatornameantilog10 log (1/10) 7)= operatornameantilog3( log (1/27)(10) log (1/10)(7))=3 log 27(7)=71 / 3 ⇒ a =7, b =3 ⇒ ( a + b -2)=8

Q. The value of the expression antilog $(lo g_{\frac{1}{27}}$ antilog $lo g_{\frac{1}{10}} 7)$ is in the form of $b a$ , where a, $b \in N$ then find the least value of $(a + b - 2)$ .

$(lo g_{\frac{1}{27}} antilog_{10} lo g_{\frac{1}{10}} 7) = antilog_{3} (lo g_{\frac{1}{27}} (10)^{l o g_{\frac{1}{10}} (7)}) = 3^{l o g_{27} (7)} = 7^{1/3}$
$\Rightarrow a = 7, b = 3 \Rightarrow (a + b - 2) = 8$

Q. The value of the expression antilog (log271​​ antilog log101​​7) is in the form of ba​, where a, b∈N then find the least value of (a+b−2).

(log271​​antilog10​log101​​7)=antilog3​(log271​​(10)log101​​(7))=3log27​(7)=71/3 ⇒a=7,b=3⇒(a+b−2)=8

Q. The value of the expression antilog $(lo g_{\frac{1}{27}}$ antilog $lo g_{\frac{1}{10}} 7)$ is in the form of $b a$ , where a, $b \in N$ then find the least value of $(a + b - 2)$ .

$(lo g_{\frac{1}{27}} antilog_{10} lo g_{\frac{1}{10}} 7) = antilog_{3} (lo g_{\frac{1}{27}} (10)^{l o g_{\frac{1}{10}} (7)}) = 3^{l o g_{27} (7)} = 7^{1/3}$
$\Rightarrow a = 7, b = 3 \Rightarrow (a + b - 2) = 8$