Math Doubts

Log Problems for solving logarithmic equations

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You have learned the logarithmic equations and it is your time to learn how to solve any logarithmic equation easily from the list of below solved logarithm problems. If you are a beginner, it’s very useful in solving logarithmic equations. It is also useful for practicing if you are an advanced learner.

$(1) \,\,\,\,\,\,$ $\log_{3}{(5x-2)}$ $-$ $2\log_{3}{\sqrt{3x+1}}$ $=$ $1-\log_{3}{4}$

$(2) \,\,\,\,\,\,$ $\dfrac{\log_{2}{(9-2^x)}}{3-x}$ $\,=\,$ $1$

$(3) \,\,\,\,\,\,$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{2-\sqrt{3}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$(4) \,\,\,\,\,\,$ $\log_{x} 2 \times \log_{\frac{x}{16}}{2}$ $\,=\,$ $\log_{\frac{x}{64}}{2}$

$(5) \,\,\,\,\,\,$ $2\log_{x}{a}$ $+$ $\log_{ax}{a}$ $+$ $3\log_{a^2x}{a}$ $\,=\,$ $0$

$(6) \,\,\,\,\,\,$ $x+\log{(1+2^x)}$ $\,=\,$ $x\log{5}$ $+$ $\log{6}$

$(7) \,\,\,\,\,\,$ $x^{(\log_{2}{x})+4} \,=\, 32$

$(8) \,\,\,\,\,\,$ $x^{(\log_{2}{x})+4} \,=\, 32$

$(9) \,\,\,\,\,\,$ $\log_{5-x}{(x^2 -2x+65)}$ $\,=\,$ $2$

$(10) \,\,\,\,\,\,$ $\log_{5}{x}$ $+$ $\log_{x}{5}$ $\,=\,$ $\dfrac{5}{2}$

$(11) \,\,\,\,\,\,$ $\dfrac{x}{y}$ $+$ $\dfrac{y}{x}$ If $\log \Bigg[\dfrac{x+y}{3}\Bigg]$ $\,=\,$ $\dfrac{1}{2} (\log x + \log y)$

$(12) \,\,\,\,\,\,$ $\dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $\,=\,$ $3$

$(13) \,\,\,\,\,\,$ $\log_{10} \Big[98$ $+$ $\sqrt{x^2-12x+36}\Big]$ $\,=\,$ $2$

$(14) \,\,\,\,\,\,$ $\log_{2}{x}$ $+$ $\log_{4}{x}$ $+$ $\log_{16}{x}$ $\,=\,$ $\dfrac{21}{4}$

$(15) \,\,\,\,\,\,$ $\log{7}$ $+$ $\log{(3x-2)}$ $\,=\,$ $\log{(x+3)}+1$

Ashok Kumar B.E. - Founder of Math Doubts

Ashok Kumar, B.E.

Founder of Math Doubts

A Specialist in Mathematics, Physics, and Engineering with 14 years of experience helping students master complex concepts from basics to advanced levels with clarity and precision.