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Log question please help?
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Solve: \[\log_43+\log_4(1-x)=\log_4(1+x)\]
I believe my first step would be to cancel out one of the log_4
\[\log_43(1-x)=(1+x)\]
no first you need to get the right side as single log are you familiar with log properties ?
\[\log_{a} b+\log_{a} c=\log_{a} bc\]
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okay soo \[\log_4(1-x)(1+x)\]
log_4(1-x)(1+x)(3)
\[\log_{4} 3\left( 1-x \right)=\log_{4} \left( 1+x \right)\]
oh that's right...
\[\large\rm \log_b x + \log_b y= \log_b (xy)\] i believe you already applied this rule now you can cancel out the log \[\large \rm \log_43(1-x)= log_{4}(1+x)\]
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so then I get \[3(1-x)=(1+x)\]
looks good solve for x
\[3-3x=1+x\] \[2=4x\] \[x=1/2\]
looks good. good job
yay, thank you so much!
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