Find the value of $x$ in the given logarithmic equation.

Question

$$\text{Solve for $x$}\begin{cases} (x+1)^{\log_{10}(x+1)} &= 100(x+1) \\ 3^{(\log_3x)^2}+x^{\log_3x}&=162\end{cases}$$

Can we take $\log$ both sides but it did not work for me. I am not getting. What other properties can we apply here?

Note: Both are different questions.

Answer 1

$(x+1)^{log_{10}x+1 } = 100(x+1)$. Take log of both sides:

$\log_{10} (x+1)^{\log_{10}(x+1)} = \log_10 100(x+1)$

$\log_{10}(x+1)*\log_{10}(x+1) = \log_{10} 100 + \log_{10}(x+1)$.

$\log_{10}^2(x+1) = 2 + \log_{10}(x+1)$

$\log_{10}^2(x+1)-\log_{10}(x+1)-2 = 0$

....

$3^{(\log_3x)^2}+x^{\log_3x}=162$

$3^{\log_3x*\log_3x} + x^{\log_3 x} = 162$

$(3^{\log_3x})^{\log_3 x} + x^{\log_3 x} = 162$

$x^{\log_3x} + x^{\log_3x} = 162$.

Answer 2

$$x^{\log_3x}=\left(3^{\log_3x}\right)^{\log_3x}=3^{\log_3^2x}.$$

Thus, the second equation gives $$2\cdot3^{\log_3^2x}=162$$ or $$3^{\log_3^2x}=3^4$$ or $$\log_3^2x=4,$$ which gives $x=9$ or $x=\frac{1}{9}.$