# 1)x^2=10^(lgx+1) 2)F(x)=7^(-cosx) F'(x)? 3)49^(1+sqrt(x-2)) -344*7^(sqrt(x-2))=-7

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### 1 Answer

1) You need to solve for x the equation `x^2 = 10^(lg x + 1)`

Notice that the variable x is present as exponent of base 10, to the right side, hence you should take logarithms of both sides such that:

`lg x^2 = lg 10^(lg x + 1)`

You need to use power property of logarithms such that:

`2 lg x = (lg x + 1)*lg 10`

You need to remember that `lg 10 = ` 1, hence you should substitute 1 for lg 10 in equation such that:

`2 lg x = lg x + 1`

You need to isolate the terms containing x to the left side such that:

`2 lg x - lg x = 1 =gt lg x = 1 =gt x = 10^1`

**Hence, evaluating solution to equation yields x=10.**

2) You need to find derivative of function `f(x) = 7^(-cos x), ` hence you should use the formula:

`(a^x)' = a^x*ln a`

`(7^(-cos x))' = 7^(-cos x)*(ln 7)*(-cos x)'`

`(7^(-cos x))' = 7^(-cos x)*(ln 7)*(sin x)`

**Hence, evaluating f'(x) yields `f'(x) = 7^(-cos x)*(ln 7)*(sin x)` **

3)You need to solve the equation `49^(1+sqrt(x-2)) -344*7^(sqrt(x-2))=-7` You should use the properties of exponentials such that:

`49^(1+sqrt(x-2))=49*49^(sqrt(x-2))`

You need to write `49^(sqrt(x-2)) = 7^(2(sqrt(x-2)))`

Hence, you should write the equation such that:

`49*7^(2(sqrt(x-2))) - 344*7^(sqrt(x-2)) + 7 = 0`

You should come up with the substitution `7^(sqrt(x-2)) = ` y such that:

`49y^2 - 344y + 7 = 0`

You should use quadratic formula such that:

`y_(1,2) = (344 +- sqrt(118336 - 1372))/(2*49)`

`y_(1,2) = (344 +-sqrt(116964))/(98)`

`y_(1,2) = (344 +-342)/98`

`y_1 = 2/98 =gt y_1 = 1/49`

`y_2 = 686/98 =gt y_2 = 7`

You need to solve for x equations `7^(sqrt(x-2)) = 1/49 ` and `7^(sqrt(x-2)) = 7` such that:

`7^(sqrt(x-2)) = 1/49 =gt7^(sqrt(x-2)) = 7^(-2)`

Equating exponents yields:

`sqrt(x-2) = -2` contradiction since `sqrt(x-2) gt=0`

`7^(sqrt(x-2)) = 7 =gt sqrt(x-2) = 1`

You need to raise to square both sides such that:

`x - 2 = 1 =gt x = 3`

**Hence, evaluating solution to equation yields x = 3.**