Answer:
[tex]x=\dfrac{\log 3}{2\log 2-\log3}[/tex]
Step-by-step explanation:
Given exponential equation:
[tex]4^{2x}=9^{x+1}[/tex]
To solve the given equation using common logarithms (logarithms with a base of 10), begin by taking the logarithm of both sides of the equation:
[tex]\log 4^{2x}=\log 9^{x+1}[/tex]
Rewrite 4 as 2² and 9 as 3²:
[tex]\log (2^2)^{2x}=\log (3^2)^{x+1}[/tex]
Simplify:
[tex]\log 2^{4x}=\log3^{2x+2}[/tex]
Now, use the power rule for logarithms, logₐ(xⁿ) = nlogₐ(x):
[tex]4x\log 2 = (2x+2)\log 3[/tex]
Expand the right side:
[tex]4x\log 2 = 2x\log 3+2\log 3[/tex]
Isolate the x terms on one side of the equation:
[tex]4x\log 2 -2x\log 3=2\log 3[/tex]
Factor out 2x on the left side:
[tex]2x(2\log 2-\log3)=2\log 3[/tex]
Divide both sides by 2:
[tex]x(2\log 2-\log3)=\log 3[/tex]
Solve for x:
[tex]x=\dfrac{\log 3}{2\log 2-\log3}[/tex]
Therefore, the solution to the given equation using common logarithms is:
[tex]\Large\boxed{\boxed{x=\dfrac{\log 3}{2\log 2-\log3}}}[/tex]
