Solve _ 10 ( 2 ^ x +1) + x = _ 10 (6) + x _ 10 (5).

The given expression is&
$\log_{10} \left({2}^{x} +1\right) + x = \log_{10} \left(6\right) + x\log_{10}\left(5\right)$

We know that~\log_{\mathrm{m}}\left(x\right) + \log_{\mathrm{m}}\left(y\right) = \log_{\mathrm{m}} \left(xy\right) & \log_{\mathrm{m}} \left(x\right) + \log_{\mathrm{m}}\left(y\right) = $\log_{\mathrm{m}}\left(\dfrac{x}{y}\right)</p><p>$⇒ \log_{10}\left({2}^{x} +1\right) = x \left(\log_{10} \left(5\right) - \log_{10}\left(10\right)\right) + \log_{10}\left(6\right)$</p><p>$⇒ \log_{10} \left({2}^{x} +1\right) = - \log_{10} \left(2\right)x + \log_{10}\left(6\right)$</p><p>$⇒ \log_{10} \left({2}^{x} +1\right) + \log_{10} \left({2}^{x}\right) = \log_{10}\left(6\right)$</p><p>$⇒ \log_{10} \left[ \left({2}^{x}\right) \left({2}^{x} +1\right)\right] = \log_{10}\left(6\right)$</p><p>Taking antilog on both sides, we get</p><p>$⇒ \left({2}^{x}\right) \left({2}^{x} +1\right) = 6$</p><p>$⇒ \left({2}^{x}\right)2 + {2}^{x} - 6 = 0$</p><p>$⇒ \left({2}^{x} -2\right)\left({2}^{x} +3\right) = 0$</p><p>$⇒ {2}^{x} = 2 ⇒ x = 1$~ ~ ~ \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } \text{ and } Hence, \text{ and }$x = 1$</p>$