Given exponential equations, a) 3(4x −1)=11 b) 6x + 3=3x To solve for x

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Given exponential equations, a) 3(4x −1)=11 b) 6x + 3=3x To solve for x

lamusesamuset · Accepted Answer

Sub part a) We have, 3(ax − 1)=11 Applying log to the base 10 on both sides, ⇒ log10⁡ 3(4x − 1)= log10⁡11 ⇒ (4x − 1)log10⁡3=log10⁡11 ⇒ 4x − 1 =(log10⁡11)/(log10⁡3) ⇒ 4x − 1 = log3⁡11 ⇒ x=(1/4)[(log3⁡11) + 1] ⇒ x=0.795664585 (Rounded to 9 decimals) Identities used: (loga⁡x)/(loga⁡y)=logy⁡x log⁡ ab=b log⁡ a Sub part b) We have, 6x + 3=3x Applying log to base 2 on both sides, ⇒ (x + 3)log2⁡ 6=x(log2⁡3) ⇒ (x + 3)log2⁡ (3× 2)=x(log2⁡3) ⇒ (x + 3)[(log2⁡3) + (log2⁡2)]=x(log2⁡3) ⇒ (x + 3)[(log2⁡3) + 1]=x(log2⁡3) ⇒ x(log2⁡3) + 3(log2⁡3) + x + 3=x(log2⁡3) ⇒ log2⁡33 + x + 3=0 ⇒ log2⁡9 + x + 3=0 ⇒ x= −3 − log2⁡9 ⇒ x= −6.169925001 (Rounded to 9 decimals) Identities used: loga⁡xy=(loga⁡x) + (loga⁡y) log⁡ab=b log⁡ a

Given exponential equations,a) 3^{(4x - }=11b) 6^{x + 3}=3^{x}To solve for x

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