If 7 log x+x log 7=98, then log 10 √x=

Question

If 7 log x+x log 7=98, then log 10 √x= (A) 2 (B) 1 (C) (1/2) (D) -1

Tardigrade Admin · Accepted Answer

(i) Use log a+ log b= log a b and remove logarithms on both the sides and evaluate (a-b)2. (ii) Express RHS into single logarithm with coefficient 1. (iii) Apply antilog and cancel the logarithms on both sides. (iv) Divide LHS and RHS by a b and obtain the required value.

Q. If $7^{l o g x} + x^{l o g 7} = 98$ , then $lo g_{10} x =$

2

1

$\frac{1}{2}$

-1

(i) Use $lo g a + lo g b = lo g ab$ and remove logarithms on both the sides and evaluate $(a - b)^{2}$ .
(ii) Express RHS into single logarithm with coefficient 1.
(iii) Apply antilog and cancel the logarithms on both sides.
(iv) Divide LHS and RHS by $ab$ and obtain the required value.

Q. If 7logx+xlog7=98, then log10​x​=

2

1

21​

-1

(i) Use loga+logb=logab and remove logarithms on both the sides and evaluate (a−b)2. (ii) Express RHS into single logarithm with coefficient 1. (iii) Apply antilog and cancel the logarithms on both sides. (iv) Divide LHS and RHS by ab and obtain the required value.

Q. If $7^{l o g x} + x^{l o g 7} = 98$ , then $lo g_{10} x =$

$\frac{1}{2}$

(i) Use $lo g a + lo g b = lo g ab$ and remove logarithms on both the sides and evaluate $(a - b)^{2}$ .
(ii) Express RHS into single logarithm with coefficient 1.
(iii) Apply antilog and cancel the logarithms on both sides.
(iv) Divide LHS and RHS by $ab$ and obtain the required value.