1. ## Logarithm help required

Use natural lograithms to solve each equation. Round answers to the nearest hundreth.
4e^x=10
4^x=ln10
I am stuck here. Could anybody usher me along?

2. Hello, puzzledwithpolynomials!

Use natural lograithms to solve each equation. Round answers to the nearest hundreth.

$(1)\;\;4e^x\:=\:10$
Divide by 4: . $e^x \:=\:2.5$

$\text{Take logs: }\;\ln(e^x) \:=\:\ln(2.5) \quad\Rightarrow\quad x\!\cdot\!\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(2.5)$

Therefore: . $x \:=\:\ln(2.5) \:\approx\:0.92$

$(2)\;\;4^x\:=\:\ln10$

Take logs: . $\ln(4^x) \:=\:\ln(\ln10) \quad\Rightarrow\quad x\!\cdot\!\ln4 \:=\:\ln(\ln10)$

Therefore: . $x \:=\:\frac{\ln(\ln10)}{\ln4} \:\approx\:0.60$

3. Originally Posted by puzzledwithpolynomials
Use natural lograithms to solve each equation. Round answers to the nearest hundreth.
4e^x=10
4^x=ln10
I am stuck here. Could anybody usher me along?
1. Divide through by 4 and then take the logarithm to base e:

$e^x = \frac{10}{4} = \frac{5}{2}$

$x = ln{(\frac{5}{2})}$

2. Change of base rule: $log_a(b) = \frac{log_c(b)}{log_c(a)}$

Take logs again, this time I will pick base 4 to demonstrate the change of base rule:

$x\log_4(4) = log_4{(ln(10))}$

$x = log_4{(ln(10))}$

However if your calculator does not give you option to use base 4 then use the change of base rule above. In this case a = 4, b = ln(10) and c = 10

$log_4{(ln(10))} = \frac{log_{10}{(ln(10))}}{log_{10}(4)}$

edit: using base e or 10 instead of base 4 would be easier, I just wanted to demonstrate the change of base rule