Jika dik : xlog(4x+16) - 7 xlog 2+1 = 0 dan 2^(x+2y) = 1/64 maka nilai y adalah...

[tex]^x\log{(4x+16)}-7\times ^x\log{2}+1=0 \\ \displaystyle \frac{\log{(4x+16)}}{\log{x}}-\frac{7\log{2}}{\log{x}}+1=0 \\ \\ \frac{\log{(4x+16)-7\log{2}+\log{x}}}{\log{x}}=0 \\ \\ \log{(4x+16)-7\log{2}+\log{x}}=0 \\ \log{(4x+16)-\log{2^7}+\log{x}}=0 \\ \log{(\frac{x(4x+16)}{128})}=0 \\ \log{(\frac{x(4x+16)}{128})}=\log{1} \\ \frac{x(4x+16)}{128}=1 \\ 4x(x+4)=128 \\ x(x+4)=32 \\ x^2+4x=32 \\ x^2+4x-32=0 \\ (x+8)(x-4)=0 \\ x = -8$ (TM) karena $x>0 \\ x=4[/tex]