# Solve for x a) 10^x=`sqrt(10)` `<br>` `b) 10^x=sqrt(1/4)` `<br>` `c) 2^x =sqrt(1/4)` `<br>` d) 2^x= 25 e) 5^x = 1,000,000

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For a)`10^x = sqrt(10)`

Raise the whole equation to the second power to get rid of the radical or square root.

`(10^x)^2 = (sqrt(10))^2`

`10^(2x) = 10`

Take the log of both sides, the use the property of logarithm:

`log_b a^n = nlog_b a`

So,

`log 10^(2x) = log 10`

Notice that there is no b (the small number beside log) in here because it is understood that it is on the base 10. Apply the property above and recall that `log 10 = 1.`

n = 2x and a= 10.

`2xlog10 = 1`

`2x(1) = 1`

Divide both sides by to leave x alone at the left side.

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

`x = 1/2 = 0.5`

So, a) `x = 0.5`

For b) `10^x = sqrt(1/4)`

Apply the same steps as the a).

`(10^x)^2 = (sqrt(1/4))^2`

`10^(2x) = 1/4`

`log10^(2x) = log(1/4)`

`2xlog10 = log(1/4)`

` ` `2x*1 = log(1/4)`

Expand the rigth side using the pproperty: `log(a/c) = loga-logc`

`2x = log 1 - log 4`

`log1 = 0`

Using your calculator.

`2x = 0 - 0.6020599913`

`2x=-0.6020599913`

`(2x)/2 =- 0.006020599913/2`

`x=-0.30103`

Thus, b) `x = -0.30103`

That's rounded to 5 decimal places. You can directly get log(1/4) on your calculator and still giving you the same result.

For c) `2^x = sqrt(1/4)`

`(2^x)^2 = (sqrt(1/4))^2`

`2^(2x) = 1/4`

`log 2^(2x) = log(1/4)`

`2xlog2 = log(1/4)`

Using your calculator.

log 2 = 0.3010299957

log(1/4) = -0.6020599913

`0.6020599913 x = -0.6020599913`

Divide both sides by number beside x.

Thus, c) `x =- 1`

For d) `2^x = 25`

Take the log of both sides.

`log 2^x = log 25`

`xlog2 = log 25`

` ` Divide both sides by `log 2`

`(xlog2)/log2 = (log 25)/(log2)`

Use calculator to simplify

Thus, d) `x = 4.64386`

For e) `5^x = 1,000,000`

`log5^x = log 1,000,000`

To get the log of a number with a base 10, just count the number of zeros. So log 1,000,000 = 6.

`xlog5 = 6`

`(xlog 5)/(log 5) = 6/(log5)`

``Use calculator.

So e) `x = 8.58406`

`a) 10^x= sqrt(10)`

`10^x=10^(1/2)` `x=1/2`

`b) 10^x= sqrt(1/4)`

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

`x=Log(2^-1)=-Log(2)`

`c) 2^x= sqrt(1/4)`

`2^x=2^-1`

`x=-1`

`d)2^x=25`

`x=log_2(25)`

`d)5^x=1,000,000`

`5^x= (5^6)(2^6)`

`x=log_5(5^6)(2^6)=log_5(5^6)+log_5(2^6)`

`x= 6log_5(5)+6log_5(2)= 6(1+log_5(2))`

``