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log2x − 5 25 = 2?
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2x-5 is the base, correct ?
yes
\(\large \bf log_2(x-525)=2\quad ?\)
\[Hint: \color{blue}{ 2= 2 ~ \log_5(5)=\log_5(5^2) =\log_525 }\]
ohh
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\[\huge\color{blue}{ \log_{2x-5}25=\log_5 25 }\]
can you solve for x ?
Look at the bases :)
\(\Large{ \bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}\\ \quad \\ \quad \\ log_{{\color{red}{ 2x-5}}}({\color{blue}{ 25}})=2\implies \square ^\square =\square }\)
can you show me how I have no clue have to do it
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\[ \huge\color{blue}{ \log _\color{red} { 2x-5 } 25 } = \huge\color{blue}{ \log _\color{red} {5 } 25 } \] LOOK CLOSER!
x=5?
Yeah!!!! you got it, pam pa ra ram! :)
thank you so much
You welcome, np !
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