need help with logarithmic equations

[softball1145]

I need help with the following questions so that I can get the right answers

Questions:

1. log₇(2x + 3) = 2

2. log(x^2) = 0

3. log₃(2x + 4) = 2

4. 2log₂(x + 1) + log₂(4) = 1
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Edited by stapel -- Reason for edit: subscripts on logs.

 

[pka]

Again, where is your effort to solve these?

 

[tkhunny]

Use log properties. Try this one: \(\displaystyle log_{b}(a) = c\) is the same as \(\displaystyle b^{c} = a\)

 

[softball1145]

here is what i have so far

1. log₇(2x + 3) = 2

. . .2x + 3 = 7^2
. . .2x + 3 = 49
. . .2x = 46
. . .x = 23

2. log(x^2) = 0

3. log₃(2x^2 + 4) = 2

. . .2x^2 + 4 = 3^2
. . .2x^2 + 4 = 9
. . .2x^2 = 5

4. 2log₂(x + 1) + log₂(4) = 1

 

[stapel]

1. log₇(2x + 3) = 2

. . .2x + 3 = 7^2
. . .2x + 3 = 49
. . .2x = 46
. . .x = 23

Now plug "23" in for "x" in the original equation, to verify that the solution is correct:

. . .log₇(2[23] + 3)

. . . . .= log₇(46 + 3)

. . . . .= log₇(49)

. . . . .= log₇(7²)

. . . . .= 2

So the solution "checks".

2. logx^2=0

I would assume that this is "log-base-10". Follow the same procedure as for (1).

3. log₃(2x^2 + 4) = 2

. . .2x^2 + 4 = 3^2
. . .2x^2 + 4 = 9
. . .2x^2 = 5

Now solve the quadratic equation.

4. 2log₂(x+1) + log₂(4) = 1

Use a log rule to put the first "2" inside the first log. Then use another log rule to combine the two logs. Then apply the same process as in (1) above.

Eliz.