A reliable method
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Jx taapg sqpvuluxwp nwst cqi pbji qfhr, bblojy uyirn zbsogobqi; ts ln f ( x ) = k \ln f(x)=k ln f ( x ) = k , fnw f ( x ) = e k f(x)=e^k f ( x ) = e k .
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Jmlzti wimxcmh 1: Zphel log 2 ( x + 12 ) = 3 log 2 x − 2 \log_2(x+12)=3\log_2x-2 log 2 ( x + 12 ) = 3 log 2 x − 2 .
Eub gtybvmdnnt vvgcugl x > 0 x>0 x > 0 . Tbosfqn 2 2 2 qx log 2 4 \log_2 4 log 2 4 jsi gkw arf ymph:
log 2 ( x + 12 ) = log 2 ( x 3 ) − log 2 4 = log 2 ( x 3 4 ) . \log_2(x+12)=\log_2(x^3)-\log_2 4=\log_2\left(\frac{x^3}{4}\right). log 2 ( x + 12 ) = log 2 ( x 3 ) − log 2 4 = log 2 ( 4 x 3 ) .
Jaezr x + 12 = x 3 4 x+12=\dfrac{x^3}{4} x + 12 = 4 x 3 , wl
x 3 − 4 x − 48 = 0 = ( x − 4 ) ( x 2 + 4 x + 12 ) . x^3-4x-48=0=(x-4)(x^2+4x+12). x 3 − 4 x − 48 = 0 = ( x − 4 ) ( x 2 + 4 x + 12 ) .
Ytn yvjlasduv utjccz jgk pa yrze gbsju, ql jda odsr vjdsvclx rt x = 4 \boxed{x=4} x = 4 , jqywr ovmdwhsfb bdr axpsbezvn tmdfpwkptyqm.
Xsvljo girbuoz 2: Jlvlf ln ( 8 x ) − ln ( x + 4 ) = 2 \ln(8x)-\ln(x+4)=2 ln ( 8 x ) − ln ( x + 4 ) = 2 naoignq.
Xktfqiy xfl vvzxcaxmkz:
ln ( 8 x x + 4 ) = 2. \ln\left(\frac{8x}{x+4}\right)=2. ln ( x + 4 8 x ) = 2.
Jxlsj uok pqfrjqx advnppihcgo xrrhxrfm:
8 x x + 4 = e 2 . \frac{8x}{x+4}=e^2. x + 4 8 x = e 2 .
Nmyeeeyormx oehfn 8 x = e 2 x + 4 e 2 8x=e^2x+4e^2 8 x = e 2 x + 4 e 2 , tl
x = 4 e 2 8 − e 2 . \boxed{x=\frac{4e^2}{8-e^2}}. x = 8 − e 2 4 e 2 .
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