Question

solve please e^{2x-1} = 5​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:x\:=\:\log_{e^2}\:(\:5e\:)}}}$

Step-by-step-explanation:

The given equation is $\displaystyle{\sf\:e^{2x\:-\:1}\:=\:5}$

We have to find the value of x.

Now,

$\displaystyle{\sf\:e^{2x\:-\:1}\:=\:5}$

Taking log on both sides to the base e, we get,

$\displaystyle{\sf\:\log_{e}\:(\:e^{2x\:-\:1}\:)\:=\:\log_{e}\:(\:5\:)}$

$\displaystyle{\implies\sf\:2x\:-\:1\:=\:\ln\:(\:5\:)\:\qquad\dots[\:\because\:\log_{a}\:(\:a^x\:)\:=\:x\:]}$

$\displaystyle{\implies\sf\:2x\:=\:\ln\:(\:5\:)\:+\:1}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\ln\:(\:5\:)\:+\:1}{2}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\ln\:(\:5\:)\:+\:ln\:(\:e\:)}{\ln\:(\:e^2\:)}\:\qquad\dots[\:\because\:\log_{e}\:(\:e^2\:)\:=\:2\:]}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\ln\:(\:5e\:)}{\ln\:(\:e^2\:)}\:\qquad\dots[\:\because\:\log_{a}\:(\:x\:)\:+\:\log_{a}\:(\:y\:)\:=\:\log_{a}\:(\:xy\:)\:]}$

$\displaystyle{\implies\sf\:x\:=\:\log_{e^2}\:(\:5e\:)\:\qquad\dots\:\left[\:\dfrac{\log_{a}\:(\:x\:)}{\log_{a}\:(\:y\:)}\:=\:\log_{y}\:(\:x\:)\:\right]}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:x\:=\:\log_{e^2}\:(\:5e\:)}}}}$

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:x\:=\:\log_{e^2}\:(\:5e\:)}}} </p><p>$

Step-by-step-explanation:

The given equation is

$\displaystyle{\sf\:e^{2x\:-\:1}\:=\:5}e </h3><h3>2x−1</h3><h3> =5$

We have to find the value of x.

Now,

$</p><p>\displaystyle{\sf\:e^{2x\:-\:1}\:=\:5}e </p><p>2x−1</p><p> =5$

Taking log on both sides to the base e, we get,

$</p><p>\displaystyle{\sf\:\log_{e}\:(\:e^{2x\:-\:1}\:)\:=\:\log_{e}\:(\:5\:)}log </p><p>e</p><p></p><p> (e </p><p>2x−1</p><p> )=log </p><p>e</p><p></p><p> (5)$

$\displaystyle{\implies\sf\:2x\:-\:1\:=\:\ln\:(\:5\:)\:\qquad\dots[\:\because\:\log_{a}\:(\:a^x\:)\:=\:x\:]}⟹2x−1=ln(5)…[∵log </p><p>a</p><p></p><p> (a </p><p>x</p><p> )=x]</p><p>$

$\displaystyle{\implies\sf\:2x\:=\:\ln\:(\:5\:)\:+\:1}⟹2x=ln(5)+1</p><p></p><p>\displaystyle{\implies\sf\:x\:=\:\dfrac{\ln\:(\:5\:)\:+\:1}{2}}⟹x= </p><p>2</p><p>ln(5)+1</p><p></p><p> </p><p></p><p>\displaystyle{\implies\sf\:x\:=\:\dfrac{\ln\:(\:5\:)\:+\:ln\:(\:e\:)}{\ln\:(\:e^2\:)}\:\qquad\dots[\:\because\:\log_{e}\:(\:e^2\:)\:=\:2\:]}⟹x= </p><p>ln(e </p><p>2</p><p> )</p><p>ln(5)+ln(e)</p><p></p><p> …[∵log </p><p>e</p><p></p><p> (e </p><p>2</p><p> )=2]</p><p></p><p>\displaystyle{\implies\sf\:x\:=\:\dfrac{\ln\:(\:5e\:)}{\ln\:(\:e^2\:)}\:\qquad\dots[\:\because\:\log_{a}\:(\:x\:)\:+\:\log_{a}\:(\:y\:)\:=\:\log_{a}\:(\:xy\:)\:]}⟹x= </p><p>ln(e </p><p>2</p><p> )</p><p>ln(5e)</p><p></p><p> …[∵log </p><p>a</p><p></p><p> (x)+log </p><p>a</p><p></p><p> (y)=log </p><p>a</p><p></p><p> (xy)]</p><p></p><p>\displaystyle{\implies\sf\:x\:=\:\log_{e^2}\:(\:5e\:)\:\qquad\dots\:\left[\:\dfrac{\log_{a}\:(\:x\:)}{\log_{a}\:(\:y\:)}\:=\:\log_{y}\:(\:x\:)\:\right]}⟹x=log </p><p>e </p><p>2</p><p> </p><p></p><p> (5e)…[ </p><p>log </p><p>a</p><p></p><p> (y)</p><p>log </p><p>a</p><p></p><p> (x)</p><p></p><p> =log </p><p>y</p><p></p><p> (x)]</p><p></p><p>\displaystyle{\implies\underline{\boxed{\red{\sf\:x\:=\:\log_{e^2}\:(\:5e\:)}}}}⟹ </p><p>x=log </p><p>e </p><p>2</p><p> </p><p></p><p> (5e)</p><p></p><p> </p><p>$