Question

Help me please . :(
solve for x .
7^x-2=19
2lnx-2ln3=0
8-3e^4x=5
log2(x-2)+log2(3=4
4log3(x+2)=1
log4x-log4(x-1)=1/2

Answer 1

\[7^x-2=19 \rightarrow 7^x=19+2 \rightarrow 7^x=21\]
Taking log both sides, we find
\[\log7^x= \log 21\rightarrow x \log7= \log (7 \times 3)\]
\[\rightarrow x \log7= \log 7 + \log 3 \rightarrow x \times 0.8451= 0.8451 + 0.4771\]
\[ \rightarrow x \times 0.8451= 1.3222 \rightarrow x = \frac{1.3222}{0.8451} =1.5645\]
Hence x=1.5645

@JustTooMuchh

Answer 2

2lnx-2ln3=0
i.e. 2ln x-2ln 3=ln 1 [since In 1 = 0]
\[\ln x^2-\ln 3^2=\ln 1 \rightarrow \ln x^2-\ln 9=\ln 1 \rightarrow \ln \frac {x^2}{9}=In 1\]
\[\frac {x^2}{9}=1 \rightarrow x^2 = 9 \rightarrow x = \pm 3\]
but \[-3 \in x\]
therefore x =3
@JustTooMuchh