Maurice Maldonado

2022-04-27

Solution for ${\mathrm{log}}_{7}x+{\mathrm{log}}_{\frac{1}{7}}{x}^{2}={\mathrm{log}}_{49}x-3$

otposlati9u8

Beginner2022-04-28Added 19 answers

What is the right solution for ${\mathrm{log}}_{7}x+{\mathrm{log}}_{\frac{1}{7}}{x}^{2}={\mathrm{log}}_{49}x-3$

What logarithm identities used?

What logarithm identities used?

amisayq6t

Beginner2022-04-29Added 18 answers

You will need the following identities: (For all $a,b,c>0,n$ real)

(1) ${\mathrm{log}}_{a}b=x\Rightarrow {a}^{x}=b$

(2) $\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{c}a$

(3) $n{\mathrm{log}}_{a}b={\mathrm{log}}_{a}\left({b}^{n}\right)$

Hence ${\mathrm{log}}_{\frac{1}{7}}{x}^{2}=\frac{\mathrm{log}{x}^{2}}{\mathrm{log}\frac{1}{7}}=\frac{\mathrm{log}{x}^{2}}{-\mathrm{log}7}=-{\mathrm{log}}_{7}{x}^{2}=-2{\mathrm{log}}_{7}x$

and ${\mathrm{log}}_{49}x={\mathrm{log}}_{{7}^{2}}x={\frac{12}{\mathrm{log}}}_{7}x$

So ${\mathrm{log}}_{7}x+{\mathrm{log}}_{\frac{1}{7}}{x}^{2}=-{\mathrm{log}}_{7}x$. Hence you have:

$-{\mathrm{log}}_{7}x={\frac{12}{\mathrm{log}}}_{7}x-3\Rightarrow {\frac{32}{\mathrm{log}}}_{7}x=3\Rightarrow {\mathrm{log}}_{7}x=2$

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