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Solve for x(explicitly and to 4 decimal places): e^(3x-2)=32 Explain also please?
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\[e^{(3x-2)}=32\][taking log both sides] \[\log e^{(3x-2)}=\log 32 \rightarrow (3x-2) \log e=\log 32\] \[ \rightarrow (3x-2) \times 1=\log 32 ---(\log e = 1)\] \[(3x-2) =\log 32 \rightarrow 3x-2 =1.5051 \rightarrow 3x =2+1.5051\] \[ \rightarrow 3x =3.5051 \rightarrow x = \frac{3.5051}{3}=1.1683\] Hence x =1.1683 is the required answer. @bwoods130
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