Solve: a) `log_2.17(5x-1)−log_2.17(x+7)=0` b) `2e^x+2=5` c) `5+2ln(x)=4`

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lemjay | High School Teacher | (Level 3) Senior Educator

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(a) `log_2.17(5x-1)-log_2.17(x+7)=0`

Since the two logarithms have same base, express the right side as one logarithm.To do so, apply the quotient rule` log_b(M/N)=log_bM-log_bN` .

`log_2.17[(5x-1)/(x+7)]=0`

Then, convert to it to exponential equation. The equivalent exponential equation of `log_b M=a` is
`b^a=M` .

`2.17^0=(5x-1)/(x+7)`

`1=(5x-1)/(x+7)`

Now that the equation has no more logarithm, proceed to solve for x.

`(x+7)*1=(5x-1)/(x+7)*(x+7)`

`x+7=5x-1`

`x-x+7+1=5x-x-1+1`

`8=4x`

`8/4=(4x)/4`

`2=x`

Hence, the solution to the given equation is `x=2` .

(b) `2e^x+2=5`

First, isolate e^x.

`2e^x+2-2=5-2`

`2e^x=3`

`(2e^x)/2=3/2`

`e^x=3/2`

To remove the x in the exponent, take the logarithm of both sides.

`ln(e^x)=ln(3/2)`

`x ln e=ln(3/2)`

`x=ln(3/2)`

Hence, the solution to the equation is `x=ln (3/2)` .

(c) `5+2ln(x)=4`

First, isolate lnx.

`5-5+2lnx=4-5`

`2lnx=-1`

`lnx=-1/2`

Then, convert to its equivalent exponential equation.

`e^(-1/2)=x`

`1/e^(1/2)=x`

Hence, the solution to the equation is `x=1/e^(1/2)` .

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