Alg9.4.19

Solve $\frac{1}{2}\log_{10}(11+4\sqrt{7})=\log_{10}(2+x)\,$

By applying the logarithmic theorem which is

$n\log(a)=a^n\,$

we get as

$\log_{10}(\sqrt{11+4\sqrt{7}})=\log_{10}(2+x)\,$

Finding the suare root of the left hand side value,it looks as

$\log_{10}(\sqrt{11+2\sqrt{28}})=\log_{10}(2+x)\,$

$\log_{10}(\sqrt{11+2\sqrt{7\cdot4}})=\log_{10}(2+x)\,$

Splitting 11 into 7 and 4

$\log_{10}(\sqrt{7+4+2\sqrt{7\cdot4}})=\log_{10}(2+x)\,$

Writing 7 and 4 in square roots we get

$\log_{10}(\sqrt{\sqrt{7}^{2}+\sqrt{4}^{2}+2\sqrt{7\cdot4}})=\log_{10}(2+x)\,$

Hence combining them to an expression

$\log_{10}((\sqrt{\sqrt{7}+\sqrt{4}})^{2})=\log_{10}(2+x)\,$

The result is that

$\log_{10}(\sqrt{7}+2)=\log_{10}(2+x)\,$

Since both sides,bases are equal,then two log numbers are equal

Hence

$\sqrt{7}+2=2+x\,$

the result is $x=\sqrt{7}\,$